Steve Jobs’ Convocation Speech (Stanford)


Thank you.
I’m honored to be with you today for your commencement from one of the finest
universities in the world. Truth be told, I never graduated from college, and this is the
closest I’ve ever gotten to a college graduation. Today, I want to tell you three stories
from my life. That’s it. No big deal. Just three stories.
The first story is about connecting the dots. I dropped out of Reed College after the
first six months, but then stayed around as a drop-in for another 18 months or so
before I really quit. So why did I drop out?
It started before I was born. My biological mother was a young, unwed graduate
student, and she decided to put me up for adoption. She felt very strongly that I
should be adopted by college graduates, so everything was all set for me to be adopted http://www.lifeofexcellence.com
at birth by a lawyer and his wife except that when I popped out they decided at the
last minute that they really wanted a girl.
So my parents, who were on a waiting list, got a call in the middle of the night asking,
“We’ve got an unexpected baby boy; Do you want him?” They said, “Of course.” My
biological mother found out later that my mother had never graduated from college
and that my father had never graduated from high school. She refused to sign the
final adoption papers. She only relented a few months later when my parents
promised that I would go to college. This was the start in my life.
And 17 years later I did go to college. But I naively chose a college that was almost as
expensive as Stanford, and all of my working-class parents’ savings were being spent
on my college tuition. After six months, I couldn’t see the value in it. I had no idea
what I wanted to do with my life and no idea how college was going to help me figure
it out. And here I was spending all of the money my parents had saved their entire
life.
So I decided to drop out and trust that it would all work out okay. It was pretty scary
at the time, but looking back it was one of the best decisions I ever made. The minute
I dropped out I could stop taking the required classes that didn’t interest me, and
begin dropping in on the ones that looked far more interesting.
It wasn’t all romantic. I didn’t have a dorm room, so I slept on the floor in friends’
rooms. I returned coke bottles for the five cent deposits to buy food with, and I would
walk the seven miles across town every Sunday night to get one good meal a week at
the Hare Krishna temple. I loved it. And much of what I stumbled into by following
my curiosity and intuition turned out to be priceless later on.
Let me give you one example:
Reed College at that time offered perhaps the best calligraphy instruction in the
country. Throughout the campus every poster, every label on every drawer, was
beautifully hand calligraphed. Because I had dropped out and didn’t have to take the
normal classes, I decided to take a calligraphy class to learn how to do this. I learned
about serif and san serif typefaces, about varying the amount of space between
different letter combinations, about what makes great typography great. It was
beautiful, historical, artistically subtle in a way that science can’t capture, and I found
it fascinating.
None of this had even a hope of any practical application in my life. But ten years
later, when we were designing the first Macintosh computer, it all came back to me. http://www.lifeofexcellence.com
And we designed it all into the Mac. It was the first computer with beautiful
typography. If I had never dropped in on that single course in college, the “Mac”
would have never had multiple typefaces or proportionally spaced fonts. And since
Windows just copied the Mac, it’s likely that no personal computer would have them.
If I had never dropped out, I would have never dropped in on that calligraphy class,
and personal computers might not have the wonderful typography that they do. Of
course it was impossible to connect the dots looking forward when I was in college.
But it was very, very clear looking backwards 10 years later.
Again, you can’t connect the dots looking forward; you can only connect them
looking backwards. So you have to trust that the dots will somehow connect in your
future. You have to trust in something your gut, destiny, life, karma, whatever
because believing that the dots will connect down the road will give you the
confidence to follow your heart, even when it leads you off the wellworn path, and
that will make all the difference.
My second story is about love and loss.
I was lucky I found what I loved to do early in life. Woz¹ and I started Apple in my
parents’ garage when I was 20. We worked hard, and in 10 years Apple had grown
from just the two of us in a garage into a two billion dollar company with over 4000
employees. We’d just released our finest creation the Macintosh a year earlier, and I
had just turned 30.
And then I got fired. How can you get fired from a company you started? Well, as
Apple grew we hired someone who I thought was very talented to run the company
with me, and for the first year or so things went well. But then our visions of the
future began to diverge and eventually we had a falling out. When we did, our Board
of Directors sided with him. And so at 30, I was out. And very publicly out. What had
been the focus of my entire adult life was gone, and it was devastating.
I really didn’t know what to do for a few months. I felt that I had let the previous
generation of entrepreneurs down –that I had dropped the baton as it was being
passed to me. I met with David Packard and Bob Noyce and tried to apologize for
screwing up so badly. I was a very public failure, and I even thought about running
away from the valley. But something slowly began to dawn on me: I still loved what I
did. The turn of events at Apple had not changed that one bit. I had been rejected, but
I was still in love. And so I decided to start over.
I didn’t see it then, but it turned out that getting fired from Apple was the best thing
that could have ever happened to me. The heaviness of being successful was replaced http://www.lifeofexcellence.com
by the lightness of being a beginner again, less sure about everything. It freed me to
enter one of the most creative periods of my life.
During the next five years, I started a company named NeXT; another company
named Pixar, and fell in love with an amazing woman who would become my wife.
Pixar went on to create the world’s first computer animated feature film, Toy Story,
and is now the most successful animation studio in the world. In a remarkable turn of
events, Apple bought NeXT, and I retuned to Apple, and the technology we
developed at NeXT is at the heart of Apple’s current renaissance. And Laurene and I
have a wonderful family together.
I’m pretty sure none of this would have happened if I hadn’t been fired from Apple. It
was awful tasting medicine, but I guess the patient needed it. Sometime life
sometimes life’s going to hit you in the head with a brick. Don’t lose faith. I’m
convinced that the only thing that kept me going was that I loved what I did. You’ve
got to find what you love.
And that is as true for work as it is for your lovers. Your work is going to fill a large
part of your life, and the only way to be truly satisfied is to do what you believe is
great work. And the only way to do great work is to love what you do. If you haven’t
found it yet, keep looking and –don’t settle. As with all matters of the heart, you’ll
know when you find it. And like any great relationship, it just gets better and better
as the years roll on. So keep looking don’t settle.
My third story is about death.
When I was 17, I read a quote that went something like: “If you live each day as if it
was your last, someday you’ll most certainly be right.” It made an impression on me,
and since then, for the past 33 years, I’ve looked in the mirror every morning and
asked myself: “If today were the last day of my life, would I want to do what I am
about to do today?” And whenever the answer has been “No” for too many days in a
row, I know I need to change something.
Remembering that I’ll be dead soon is the most important tool I’ve ever encountered
to help me make the big choices in life. Because almost everything all external
expectations, all pride, all fear of embarrassment or failure these things just fall away
in the face of death, leaving only what is truly important. Remembering that you are
going to die is the best way I know to avoid the trap of thinking you have something
to lose. You are already naked. There is no reason not to follow your heart. http://www.lifeofexcellence.com
About a year ago I was diagnosed with cancer. I had a scan at 7:30 in the morning,
and it clearly showed a tumor on my pancreas. I didn’t even know what a pancreas
was. The doctors told me this was almost certainly a type of cancer that is incurable,
and that I should expect to live no longer than three to six months. My doctor advised
me to go home and get my affairs in order, which is doctor’s code for “prepare to die.”
It means to try and tell your kids everything you thought you’d have the next 10
years to tell them in just a few months. It means to make sure everything is buttoned
up so that it will be as easy as possible for your family. It means to say your goodbyes.
I lived with that diagnosis all day. Later that evening I had a biopsy, where they stuck
an endoscope down my throat, through my stomach into my intestines, put a needle
into my pancreas and got a few cells from the tumor. I was sedated, but my wife, who
was there, told me that when they viewed the cells under a microscope the doctors
started crying because it turned out to be a very rare form of pancreatic cancer that is
curable with surgery. I had the surgery and, thankfully, I’m fine now.
This was the closest I’ve been to facing death, and I hope it’s the closest I get for a few
more decades. Having lived through it, I can now say this to you with a bit more
certainty than when death was a useful but purely intellectual concept: No one wants
to die.
Even people who want to go to heaven don’t want to die to get there. And yet death
is the destination we all share. No one has ever escaped it. And that is as it should be,
because Death is very likely the single best invention of Life. It’s Life’s change agent.
It clears out the old to make way for the new. Right now the new is you, but someday
not too long from now, you will gradually become the old and be cleared away. Sorry
to be so dramatic, but it’s quite true.
Your time is limited, so don’t waste it living someone else’s life. Don’t be trapped by
dogma which is living with the results of other people’s thinking. Don’t let the noise
of others’ opinions drown out your own inner voice.
And most important, have the courage to follow your heart and intuition. They
somehow already know what you truly want to become. Everything else is secondary.
When I was young, there was an amazing publication called The Whole Earth
Catalogue, which was one of the “bibles” of my generation. It was created by a fellow
named Stewart Brand not far from here in Menlo Park, and he brought it to life with
his poetic touch. This was in the late 60s, before personal computers and desktop
publishing, so it was all made with typewriters, scissors, and Polaroid cameras. It was http://www.lifeofexcellence.com
sort of like Google in paperback form, 35 years before Google came along. It was
idealistic, overflowing with neat tools and great notions.
Stewart and his team put out several issues of The Whole Earth Catalog, and then
when it had run its course, they put out a final issue. It was the mid1970s, and I was
your age. On the back cover of their final issue was a photograph of an early morning
country road, the kind you might find yourself hitchhiking on if you were so
adventurous. Beneath it were the words:
“Stay Hungry. Stay Foolish.” It was their farewell message as they signed off. Stay
Hungry. Stay Foolish. And I’ve always wished that for myself. And now, as you
graduate to begin a new, I wish that for you.
Stay Hungry.
Stay Foolish.
Thank you all very much.

CAT 2010 Exam Analysis


CAT 2010 – Exam Analysis

This year CAT has varied expectations. With a longer time frame, and one year to put their act together, Prometric and IIMs have trained staff, got their processes in place, and are much more organised this time. Glitches seem to be fewer.

To put you minds at ease and answer all your queries about the CBT CAT 2010, here is a short writeup which enumerates the various points that you need to keep in mind while attempting the exam.

Tutorial (optional): Duration – 15 minutes.
We suggest, you should go through the tutorial once, even if you have seen it before.

Test Duration:
The actual duration of the test is 2 hours and 15 minutes.

Sections covered:
Verbal Ability, Quantitative Ability & Data Interpretation & Logical Reasoning; not necessarily in that order.

Test Format and pattern:
As per the CAT website, there are a total of 60 questions with 3 sections.

Negative Marking:
The raw scores will be calculated with +3 marks for every correct answer and –1 for every wrong answer. This is mentioned on the CAT website.

Highlighting Text:
You can highlight text in group questions like RCs. You can make multiple highlights in a particular screen, which gets deleted once you switch to the next screen.

Reviewing Questions:
You can mark a question to come back to it later. You can then Review either all the questions or review only the marked questions or review only the incomplete questions. The review shows a list of all the questions, but it will not directly give you a count of the questions attempted or to be attempted. You will need to manually count the number of questions for such data.

Items provided for rough work:
Pencils, eraser and a paper booklet of 8 A4 size papers will be provided. You will not be allowed to take your own stationary into the computer lab. In case of broken or blunt pencil points, your pencil will be replaced immediately.

Timer:
The timer is present on the screen at all time and shows the time remaining till the end of the test. Wrist watches are not allowed into the computer lab. At the end of the test, when your time runs out, you will be automatically logged out of the test and a screen with “You have run out of time” text will be visible.

WHAT WE DON’T KNOW
1. How raw scores will be scaled and how percentiles will be calculated.

WHAT WE DO KNOW
Biometrics Testing includes your photo being taken by a PC camera and an electronic finger printing (left and right hand index finger 3-3 times).
Voucher, Admit card printout and Photo ID are required to give the test. If you have lost either, call on the Prometric helpline immediately. .
Having different addresses in the admit card and in the photo-id won’t deter you from giving the test. To be on the safe side, if such a case rises, take your electricity or telephone bill with you.
You do not require the Prometric username and password to give the test.
The password required to login to the test will be given by Prometric volunteers at the start of the test.
For doing rough work, pencil, eraser and a booklet having 4 A4 sizes papers (8 sides) are provided. No sharpners are provided. They just change your pencils.
Skipping the 15 minutes tutorial won’t grant extra time for the test. It is advisable to go through the tutorial anyway.
System is user friendly. However there are reports that questions took time to load in some centres.
Changing resolution of the screen (zooming) is not possible.
Exiting the test software or opening another application like Notepad of MS Word is not possible.
Cntrl + F is disabled. So finding a word in RC using the same is not possible.
Negative marking exists.
To clear the clicked option, re-click on the same to deselect the option.
You can jump between questions by using the Review feature (as shown in the CAT demo Player)
Highlighting multiple lines in an RC section is possible. However, the selection is lost if you move away from the section.
Time left is displayed on screen. Timer logs you out automatically at the end of the test.
You wouldn’t get a score-sheet immediately after the exam is over.
STRATEGIES
As scaling system is unknown, the only strategy in hand is to attempt as many questions as possible without compromising on accuracy.
Students claim that the level of difficulty of CAT was similar to CAT 2009. Hence attempting more questions correctly will give you an edge.
It is recommended that a break up of 40 minutes for Verbal, 45 minutes for Quant and 50 minutes for DI LR is optimum.

life


Two roads diverged in a yellow wood, And sorry I could not travel both And be one traveler, long I stood And looked down one as far as I could To where it bent in the undergrowth; Then took the other, as just as fair, And having perhaps the better claim, Because it was grassy and wanted wear; Though as for that the passing there Had worn them really about the same, And both that morning equally lay In leaves no step had trodden black. Oh, I kept the first for another day! Yet knowing how way leads on to way, I doubted if I should ever come back. I shall be telling this with a sigh Somewhere ages and ages hence: Two roads diverged in a wood, and I— I took the one less traveled by, And that has made all the difference. ~Robert Frost~

inspration


Of late, I have been reading a lot about 25 random things about people. I told myself that I wouldn’t join the bandwagon until two of my best friends compelled me to write. One made it sound like it was my duty by saying I ‘should’ write and the other ordered me saying ‘you better write it by EOD tomorrow.’ This is what you call peer pressure. So, here they are, in no order of importance .

1. For one of my first singing performances in Sunday School, I reached the church and found out that I wore different pairs of shoes. White on the left foot and black on the right! (Note this: ‘found’ because I didn’t know until someone literally pointed it out to me.)
2. I don’t like receiving forwarded text messages, especially jokes, after 10:30P.M. Arrrrrrgh!
3. I don’t like people borrowing my stuff and either not returning it, or delay in returning it.
4. I never, ever, get lonely when I am alone but I easily get lonely in a crowd/group.
5. I dream of making a living by singing.
6. I easily get attracted to people who are musically talented.
7. I traveled to Afghanistan and stayed in Kabul for 3 weeks. I shot two documentaries which was aired on ABC. The documentaries made more than a million dollar for the production house I worked for. (I got about $35, yes $35, as per dium). No regrets. I didn’t work for money. It was for a social cause. I wish to go back to Afghanistan for social work.
8. People wearing shoes without removing the price tag on the sole makes me really uncomfortable. Trust me, I notice even if the tag is really old and muddy, or even if it’s on a flat sandal.
9. I don’t listen to music when I am sad or homesick. It makes me sadder.
10. I have 4 nieces in a row, and I love them to death.
11. I am terribly, allergic to air conditioners yet I spend most of my waking hours in AC rooms. (I survive on anti-allergy tablets)
12. My favorite hymn is ‘It is well with my soul.’ I hope someone will sing it for me on my funeral.
13. I have crossed my silver year. I still intent to learn the keyboard. I don’t have a keyboard yet.
14. Thoughtless, selfish people never cease to amaze me.
15. I have 100% respect for intelligent, rich, powerful people who has 0% arrogance or pride. Humility never cease to impress me.
16. I can listen to someone play the piano the whole day. (By whole day, I mean 24hrs. After 24hrs, I am sure to fall asleep, so I am putting the cap at that.)
17. I have ZERO tolerance level for: waiting for people, ill fitted denim pants, service messages from Vodafone, power play, opportunism. SUB-ZERO tolerance level for: eve-teasing, lewdness, child abuse.
18. I can lose 3 kilos in a week by eating healthy (working out do not count) but takes more than a month of hogging on junk food to put on 3 kilos. People think it’s a boon, and I think it’s a ban.
19. I prefer news channels to movie channels.
20. I first look at a person’s footwear even before looking at the face.
21. I prefer fully manual cameras to digital cameras. I used to spend every penny of my pocket money buying film roles and developing them. It means even skipping the afternoon snack. I have enormous respect for photographers who do not tweak their pictures digitally during or after shoot. Fully manual camera/ photography/photographs will eternally be respected!
22. I am horrible with numbers and drawing. I need my fingers to calculate. I can’t draw an apple to save my life. I can’t even properly draw stick figures. No exaggeration here. My 4 year old niece is better than me. It surprises me too that I am SO bad in drawing.
23. I have over three digit pairs of shoes. They range from Rs. 90 to almost five digit figures.
24. When was in school, I used to be one of the tallest students in the class. I used to complain about my height. My grandfather told me that if I kept complaining about God’s given design/gift, He might take it back. My grandfather was right, since then, I grew only 2 inches taller.
25. I truly believe in the Sunday School song ‘Make new friends, but keep the old. One is silver and the other is gold.’
I stay in a city where mall culture is at its peak. Fortunately or unfortunately, I live a stone’s throw away from the biggest shopping malls in the city. Besides doing the obvious in malls, I also like to sit at one of the coffee corners and observe people shopping crazily, while sipping my cup of cappuccino. Whether it is ‘pay by cash’ or ‘swiping’ the addictive plastic card, over the years, the spending power of people has increased drastically.
Each time I go to my hometown, I see an obvious change. Most people are becoming richer, and the rich are richer. Every street has a new house and almost every house a new car.
Recently, when I was home, a friend asked me, ‘Our State is also becoming very advanced, right?’ As much as I disliked being a wet blanket on his happy thoughts, I also felt a strong need to tell him my honest opinion. I replied in an almost sarcastic tone, ‘I see a lot of personal development but not much of State development.’ The pace of personal development and State development is hugely unequal. Usually, as soon as you walk out of a private compound, your next step is sure to be on a puddle or on a heap of dust. Also, every second or third topic of conversation is about acquiring/buying land, a better car, constructing a new/another house, and the like, all about how and the desire to have more of everything.
I am not against comfortable lifestyle, riches or money. I believe in the gift of wealth. The Bible says ‘As for every man to whom God has given riches and wealth, and given him power to eat of it, to receive his heritage and rejoice in his labor–this is the gift of God.” (Ecclesiastes 5:19.) However, the sad reality is, there is an increase of materialism. Merriam Webster defines materialism as ‘the preoccupation with material things rather than intellectual or spiritual things.’ This is not to say, as believers, we cannot have material things. The problem arises when we develop an obsession with acquiring and caring for ‘stuff.’ Any preoccupation, obsession or fascination with anything other than God is sinful and is displeasing to God. We are to ‘love the Lord, your God, with all your heart, and with all your soul, and with all your might.’ (Deuteronomy 6:5) Therefore, God is the only One we can, and should, habitually occupy ourselves with. He alone is worthy of our complete attention, love and service. To offer these things to anything or anyone else is idolatry.
When we concern ourselves with the material world, we are easily drawn in by the ‘deceitfulness of wealth.’ We think we will be happier, more fulfilled or content, if only we had more of whatever it is we are chasing. Yes, the devil wants us to be chasing after something he knows will never satisfy us so we will be kept from pursuing that which is the only thing that can satisfy—God Himself. Materialism is the exact opposite of contentment. It causes us to strive for more, and more, all the while telling us that this will be the answer to all our needs and dreams.

Of late, I have been reading a lot about 25 random things about people. I told myself that I wouldn’t join the bandwagon until two of my best friends compelled me to write. One made it sound like it was my duty by saying I ‘should’ write and the other ordered me saying ‘you better write it by EOD tomorrow.’ This is what you call peer pressure. So, here they are, in no order of importance .

1. For one of my first singing performances in Sunday School, I reached the church and found out that I wore different pairs of shoes. White on the left foot and black on the right! (Note this: ‘found’ because I didn’t know until someone literally pointed it out to me.)
2. I don’t like receiving forwarded text messages, especially jokes, after 10:30P.M. Arrrrrrgh!
3. I don’t like people borrowing my stuff and either not returning it, or delay in returning it.
4. I never, ever, get lonely when I am alone but I easily get lonely in a crowd/group.
5. I dream of making a living by singing.
6. I easily get attracted to people who are musically talented.
7. I traveled to Afghanistan and stayed in Kabul for 3 weeks. I shot two documentaries which was aired on ABC. The documentaries made more than a million dollar for the production house I worked for. (I got about $35, yes $35, as per dium). No regrets. I didn’t work for money. It was for a social cause. I wish to go back to Afghanistan for social work.
8. People wearing shoes without removing the price tag on the sole makes me really uncomfortable. Trust me, I notice even if the tag is really old and muddy, or even if it’s on a flat sandal.
9. I don’t listen to music when I am sad or homesick. It makes me sadder.
10. I have 4 nieces in a row, and I love them to death.
11. I am terribly, allergic to air conditioners yet I spend most of my waking hours in AC rooms. (I survive on anti-allergy tablets)
12. My favorite hymn is ‘It is well with my soul.’ I hope someone will sing it for me on my funeral.
13. I have crossed my silver year. I still intent to learn the keyboard. I don’t have a keyboard yet.
14. Thoughtless, selfish people never cease to amaze me.
15. I have 100% respect for intelligent, rich, powerful people who has 0% arrogance or pride. Humility never cease to impress me.
16. I can listen to someone play the piano the whole day. (By whole day, I mean 24hrs. After 24hrs, I am sure to fall asleep, so I am putting the cap at that.)
17. I have ZERO tolerance level for: waiting for people, ill fitted denim pants, service messages from Vodafone, power play, opportunism. SUB-ZERO tolerance level for: eve-teasing, lewdness, child abuse.
18. I can lose 3 kilos in a week by eating healthy (working out do not count) but takes more than a month of hogging on junk food to put on 3 kilos. People think it’s a boon, and I think it’s a ban.
19. I prefer news channels to movie channels.
20. I first look at a person’s footwear even before looking at the face.
21. I prefer fully manual cameras to digital cameras. I used to spend every penny of my pocket money buying film roles and developing them. It means even skipping the afternoon snack. I have enormous respect for photographers who do not tweak their pictures digitally during or after shoot. Fully manual camera/ photography/photographs will eternally be respected!
22. I am horrible with numbers and drawing. I need my fingers to calculate. I can’t draw an apple to save my life. I can’t even properly draw stick figures. No exaggeration here. My 4 year old niece is better than me. It surprises me too that I am SO bad in drawing.
23. I have over three digit pairs of shoes. They range from Rs. 90 to almost five digit figures.
24. When was in school, I used to be one of the tallest students in the class. I used to complain about my height. My grandfather told me that if I kept complaining about God’s given design/gift, He might take it back. My grandfather was right, since then, I grew only 2 inches taller.
25. I truly believe in the Sunday School song ‘Make new friends, but keep the old. One is silver and the other is gold.’
I stay in a city where mall culture is at its peak. Fortunately or unfortunately, I live a stone’s throw away from the biggest shopping malls in the city. Besides doing the obvious in malls, I also like to sit at one of the coffee corners and observe people shopping crazily, while sipping my cup of cappuccino. Whether it is ‘pay by cash’ or ‘swiping’ the addictive plastic card, over the years, the spending power of people has increased drastically.
Each time I go to my hometown, I see an obvious change. Most people are becoming richer, and the rich are richer. Every street has a new house and almost every house a new car.
Recently, when I was home, a friend asked me, ‘Our State is also becoming very advanced, right?’ As much as I disliked being a wet blanket on his happy thoughts, I also felt a strong need to tell him my honest opinion. I replied in an almost sarcastic tone, ‘I see a lot of personal development but not much of State development.’ The pace of personal development and State development is hugely unequal. Usually, as soon as you walk out of a private compound, your next step is sure to be on a puddle or on a heap of dust. Also, every second or third topic of conversation is about acquiring/buying land, a better car, constructing a new/another house, and the like, all about how and the desire to have more of everything.
I am not against comfortable lifestyle, riches or money. I believe in the gift of wealth. The Bible says ‘As for every man to whom God has given riches and wealth, and given him power to eat of it, to receive his heritage and rejoice in his labor–this is the gift of God.” (Ecclesiastes 5:19.) However, the sad reality is, there is an increase of materialism. Merriam Webster defines materialism as ‘the preoccupation with material things rather than intellectual or spiritual things.’ This is not to say, as believers, we cannot have material things. The problem arises when we develop an obsession with acquiring and caring for ‘stuff.’ Any preoccupation, obsession or fascination with anything other than God is sinful and is displeasing to God. We are to ‘love the Lord, your God, with all your heart, and with all your soul, and with all your might.’ (Deuteronomy 6:5) Therefore, God is the only One we can, and should, habitually occupy ourselves with. He alone is worthy of our complete attention, love and service. To offer these things to anything or anyone else is idolatry.
When we concern ourselves with the material world, we are easily drawn in by the ‘deceitfulness of wealth.’ We think we will be happier, more fulfilled or content, if only we had more of whatever it is we are chasing. Yes, the devil wants us to be chasing after something he knows will never satisfy us so we will be kept from pursuing that which is the only thing that can satisfy—God Himself. Materialism is the exact opposite of contentment. It causes us to strive for more, and more, all the while telling us that this will be the answer to all our needs and dreams.

number system


algebra


Theory of Equations..
GENERAL EQUATION OF Nth DEGREE
Let polynomial f(x) = a0xn + a1xn – 1 + a2xn – 2 + … + an. where a0, a1, a2, ..an are rational numbers and n ³ 0. Then the values of x for which f(x) reduces to zero are called root of the equation f(x) = 0. The highest whole number power of x is called the degree of the equation.
For example
x4 – 3×3 + 4×2 + x + 1 = 0 is an equation with degree four.
x5 – 6×4 + 3×2 + 1 = 0 is an equation with degree five.
ax + b = 0 is called the linear equation.
ax2 + bx + c = 0 is called the quadratic equation.
ax3 + bx2 + cx + d = 0 is called the cubic equation.
Properties of equations and their roots
• Every equation of the nth degree has exactly n roots.
For example, the equation x3 + 4×2 + 1 = 0 has 3 roots,
The equation x5 – x + 2 = 0 has 5 roots, and so on…

• In an equation with real coefficients imaginary roots occur in pairs i.e. if a + ib is a root of the equation f(x) = 0, then a – ib will also be a root of the same equation. For example, if 2 + 3i is a root of equation f(x) = 0, 2 – 3i is also a root.

• If the coefficients of an equation are all positive then the equation has no positive root. Hence, the equation 2×4 + 3×2 + 5x + 1 = 0 has no positive root.
• If the coefficients of even powers of x are all of one sign, and the coefficients of the odd powers are all of opposite sign, then the equation has no negative root. Hence, the equation 6×4 – 11×3 + 5×2 – 2x + 1 = 0 has no negative root
• If the equation contains only even powers of x and the coefficients are all of the same sign, the equation has no real root. Hence, the equation 4×4 + 5×2 + 2 = 0 has no real root.
• If the equation contains only odd powers of x, and the coefficients are all of the same sign, the equation has no real root except x = 0. Hence, the equation 5×5 + 4×3 + x = 0 has only one real root at x = 0.
• Descartes’ Rule of Signs : An equation f(x) = 0 cannot have more positive roots than there are changes of sign in f(x), and cannot have more negative roots than there changes of sign in f( – x). Thus the equation x4 + 7×3 − 4×2 − x – 7 = 0 has one positive root because there is only change in sign. f( – x) = x4 − 7×3 − 4×2 + x – 7 = 0 hence the number of negative real roots will be either 1 or 3.

EXAMPLES:

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I have not written for a while for my CAT 2008 students, not because I didn’t want to but because the explosive growth of Totalgadha.com and TathaGat kept me busy in multidimensional roles. Gone are the long evening walks, leisurely problem-solving sessions and book readings. Now my time is spent in hurried classroom sessions, editing worksheets, answering incessant phone calls, and managing a rapidly-growing team. The pleasures and pains of entrepreneurship are numerous. Not that I am complaining. It feels good to see that an idea which was once a passing thought has blossomed into something real. It is heartening to see that you can do a lot of good in this world. TG.com was once a thought. Now it is a reality. It would give rise to more realities. Tomorrow, there would be something better than TG.com. There would be better thoughts. And in the end, it would be the students who would benefit the most by the new realities. It is reason enough to rejoice, pain of entrepreneurship notwithstanding.
One more CAT is approaching fast. There is already a tightening in my stomach, a lump in my throat. If I could have my wish, I would want all my students and TGites to clear CAT. But this is not possible, and in the end, life wouldn’t be interesting without a good mix of victories and defeats. No matter what happens on the D-day, I hope that the students don’t stop learning at least. Amen! So here is it ladies and gentlemen, a fruit of painful labour.

Reply

I have not written for a while for my CAT 2008 students, not because I didn’t want to but because the explosive growth of Totalgadha.com and TathaGat kept me busy in multidimensional roles. Gone are the long evening walks, leisurely problem-solving sessions and book readings. Now my time is spent in hurried classroom sessions, editing worksheets, answering incessant phone calls, and managing a rapidly-growing team. The pleasures and pains of entrepreneurship are numerous. Not that I am complaining. It feels good to see that an idea which was once a passing thought has blossomed into something real. It is heartening to see that you can do a lot of good in this world. TG.com was once a thought. Now it is a reality. It would give rise to more realities. Tomorrow, there would be something better than TG.com. There would be better thoughts. And in the end, it would be the students who would benefit the most by the new realities. It is reason enough to rejoice, pain of entrepreneurship notwithstanding.
One more CAT is approaching fast. There is already a tightening in my stomach, a lump in my throat. If I could have my wish, I would want all my students and TGites to clear CAT. But this is not possible, and in the end, life wouldn’t be interesting without a good mix of victories and defeats. No matter what happens on the D-day, I hope that the students don’t stop learning at least. Amen! So here is it ladies and gentlemen, a fruit of painful labour.

Reply

Re: Quadratic Equation
Venn Diagrams- Basics, Problems, Maxima and Minima

For all the CAT aspirants taking CAT in 2007 or 2008, this chapter should provide some insight into Venn diagrams and methods for solving the problems. This chapter comes on the demand of some high octane TG users who are responsible for my lack of sleep and excessive intake of caffeine last night. I hope this resolves many of their problems in Venn diagrams.

Venn diagrams are pictorial representations used to display mathematical or logical relationships between two or more given sets (groups of things). The drawing consists of two or more circles, each representing a specific group. Each Venn diagram begins with a rectangle representing the universal set. Then each set in the problem is represented by a circle. Any values that belong to more than one set will be placed in the sections where the circles overlap. A typical venn diagram is shown in the figure below:

In the figure, set A contains the multiples of 2 which are less than 30 and set B contains multiples of 3 which are less than 25. Therefore, A = {2, 4, 6, 8, 10, 12 … 26, 28} and B = {3, 6, 9, 12 … 21, 24}. The various areas in the above diagram depict the following relationships:
• Intersection (A∩B)- Denotes the set of elements that are shared by two or more given sets. In the figure
given below, the intersection of the two sets is shown.

A ∩ B = {6, 12, 18, 24}
• ‘Only A’ or ‘Only B’- The part of set A, or set B, which is not shared by any other set is known as ‘only A,’ or ‘only B.’ In the figure given below, the two parts are shown:

Only A = {2, 4, 8, 10, 14, 16, 20, 22, 26, 28}, only B = {3, 9, 15, 21}
• Union (AUB)- Denotes all the elements of the given sets taken once.

A U B = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 26, 28}
It can be seen that

The venn diagram for three sets is shown below:

It can be shown that

Problem-solving through Venn diagrams:
I use the following method to solve problems through Venn diagrams:

Solved Examples:
Of all the users on Totalgadha.com, 80% spend time in CAT Quant-DI forum whereas 60% spend time in CAT verbal forum. If only those users will crack CAT who spend time in both the forums, what percentage of users of TotalGadha
• will crack CAT?
• will not crack CAT?
Answer: n(AUB) = n(A) + n(B)  n(A∩B)  100% = 80% + 60%  n(A∩B)  n(A∩B) = 40%
Therefore, 40% users of TG will crack CAT. And 60% of users (only A + only B) will not crack CAT.
NOTE: See that the surplus (superfluous part) can only be adjusted inside the area denoted for the intersection of the sets, a fact we will use in maxima- minima type of questions.
A survey on a sample of 25 new cars being sold at a local auto dealer was conducted to see which of the three popular options — air conditioning, radio and power windows — were already installed. The survey found:
15 had air conditioning
2 had air conditioning and power windows but no radios
12 had radio
6 had air conditioning and radio but no power windows
11 had power windows
4 had radio and power windows
3 had all three options.
What is the number of cars that had none of the options? (CAT 2003)
1. 4 2. 3 3. 1 4. 2
Answer: We make the Venn diagram and start filling the areas as shown:

Total Number of cars according to the diagram = 2 + 6 + 3 + 1 + 5 + 2 + 4 = 23.
Therefore, number of cars having none of the given options = 25  23 = 2.
New Age Consultants have three consultants Gyani, Medha and Buddhi. The sum of the number of projects handled by Gyani and Buddhi individually is equal to the number of projects in which Medha is involved. All three consultants are involved together in 6 projects. Gyani works with Medha in 14 projects. Buddhi has 2 projects with Medha but without Gyani, and 3 projects with Gyani but without Medha. The total number of projects for New Age Consultants is one less than twice the number of projects in which more than one consultant is involved. (CAT 2003- Leaked)
What is the number of projects in which Gyani alone is involved?
1. 0
2. 1.
3. 4.
4. cannot be determined
What is the number of projects in which Medha alone is involved?
1. 0
2. 1.
3. 4.
4. cannot be determined
Answer: The Venn diagram for the three consultants is shown below:

Total Number of projects = 2  number of projects in which more than one consultant is involved  1 = 2  19  1 = 37.
Therefore, X + 8 + 6 + 3 + Y + 2 + X + Y  16 = 37  X + Y = 17. The values of X or Y cannot be uniquely determined. Medha alone is involved in X + Y  16 = 17  16 = 1 project.
Concept of Maxima and Minima:
1. When the total number of elements is fixed
Let’s have a look at the Venn diagram of two sets again:

Imagine that in the beginning, the number of elements in all the areas is zero, as shown above. All the sets are empty right now.
Let’s see what happens if I insert one element inside A∩B:

We can see that adding 1 element to A∩B increases the number of elements in both A and B by 1. The total number of elements in all areas combined is 1 only (0 + 1 + 0) but if you add the number of elements in A and B (A + B), the addition will come up to 2. Therefore, adding 1 element to A∩B gives an extra 1 element. Hence, for every surplus of 1 element we can add 1 element to A∩B.
Let’s see the Venn diagram for 3 sets:

In diagram 1, we have added 1 element to intersection of only two sets (A and B but not C). We can see that A and B both increase by 1 and therefore we get a surplus of 1 element.
In diagram 2, we have added 1 element to intersection of all the three sets (A and B and C). We can see that A, B and C all three increase by 1 element each and therefore we get a surplus of 2 elements.
Therefore, in case of three sets, we can accommodate the surplus by
• adding elements to intersection of only two sets in which case a surplus of 1 element can be accommodated by increase of 1 element in the intersection of only two sets.
• adding elements to intersection of three sets in which case a surplus of 2 elements can be accommodated by increase of 1 element in the intersection of three sets.
How is this related to maxima and minima?
Let’s see:
According to a survey, at least 70% of people like apples, at least 75% like bananas and at least 80% like cherries. What is the minimum percentage of people who like all three?
Answer: Let’s first calculate the surplus:
percentage of people who like apples + percentage of people who like bananas + percentage of people who like cherries = 70% + 75% + 80% = 225%  a surplus of 125%.
Now this surplus can be accommodated by adding elements to either intersection of only two sets or to intersection of only three sets. As the intersection of only two sets can accommodate only a surplus of 100%, the surplus of 25% will still be left. This surplus of 25% can be accommodated by adding elements to intersection of three sets. For that we have to take 25% out of the intersection of only two sets and add it to intersection of three sets. Therefore, the minimum percentage of people who like all three = 25%
The question can be solved mathematically also. Let the elements added to intersection of only two sets and intersection of three sets be x and y, respectively. These elements will have to cover the surplus.
x + 2y = 125%, where x + y  100%. For minimum value of y, we need maximum value of x.
 x = 75%, y = 25%.
In a college, where every student follows at least one of the three activities- drama, sports, or arts- 65% follow drama, 86% follow sports, and 57% follow arts. What can be the maximum and minimum percentage of students who follow
• all three activities
• exactly two activities
Answer: Let us again see the surplus:
Percentage of students who follow drama + Percentage of students who follow sports + Percentage of students who follow arts = 65% + 86% + 57% = 208%  surplus = 108%. This surplus can be accommodated through adding elements either to intersection of only two sets or to intersection of only three sets. As the intersection of only two sets can accommodate only a surplus of 100%, the surplus of 8% will still be left. This surplus of 8% can be accommodated by adding elements to intersection of three sets. For that we have to take 8% out of the intersection of only two sets and add it to intersection of three sets. Therefore, the minimum percentage of people who like all three = 8%. In this case the percentage of students who follow exactly two activities will be maximum = 92%.
The surplus of 108% can also be accommodated through adding elements to only intersection of three sets. As adding 1 element to intersection of three sets give a surplus of 2 sets, adding 54% to intersection of three sets will give a surplus of 108%. Therefore, the maximum value of students who follow all three activities is 54%. In this case the percentage of students who follow exactly two activities will be minimum = 0%.
We can also solve it mathematically  x + 2y = 108%, where x + y  100%. The maximum value of x will give minimum value of y, whereas minimum value of x will give maximum value of y.
2. When the total number of elements is NOT fixed
In this case we assign the variables to every area of the Venn diagram and form the conditions keeping two things in mind:
• try to express the areas in the Venn diagram through least number of variables.
• all the numbers will be zero or positive. No number can be negative.
Out of 210 interviews of IIM- Ahmedabad, 105 CAT crackers were offered tea by the interview panel, 50 were offered biscuits, and 56 were offered toffees. 32 CAT crackers were offered tea and biscuits, 30 were offered biscuits and toffees, and 45 were offered toffees and tea. What is the
• maximum and minimum number of CAT crackers who were offered all three snacks?
• maximum and minimum number of CAT crackers who were offered at least one snack?
Answer: Let’s make the Venn diagram for this question. Since we want to assume least number of variables, we can see that assuming a variable for the number of students who were offered all three snacks will help us express all the other areas. Let the number of students who were offered all three snacks = x.

In the above diagram, we have expressed all the areas in terms of x. To decide maximum value of x, we note that 32  x, 45  x and 30  x will be zero or positive. Therefore, the maximum value of x will be 30. (30 is the lowest among 30, 32 and 45). To decide minimum value of x, we note than x  19 and x – 12 will be zero or positive. Therefore, x cannot be less than 19 (19 is the higher number between 19 and 12).
Therefore, maximum and minimum number of CAT crackers who were offered all three snacks = 30 and 19.

The number of CAT crackers who were offered at least one snack = Total number of CAT crackers in the Venn diagram = x + 28 + 32  x + x + 45  x + x  19 + 30  x + x  12 = 104 + x.
As the maximum and minimum values of x are 30 and 19, respectively, the maximum and minimum value of 104 + x will be 134 and 123, respectively.
Maximum and minimum number of CAT crackers who were offered at least one snack = 134 and 123.
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Although the questions on progressions may not come directly in MBA exams, the theory behind progressions is used in every place where we need to sum up numbers. During your CAT 2009 preparations, the formula used in progressions should become second nature to you as they will save a lot of time. Also, the methods of summing up various types of progressions, arithmetic, geometric or otherwise, should be very clear to you so that you are able to instantly spot the type of series you are facing. Knowing the basic methods of progressions also helps you simplify a lot of complex series. So let’s start with some basic progressions and their properties:
Arithmetic Progression
Numbers are said to be in Arithmetic Progression (A.P.) when the difference between any two consecutive numbers in the progression is constant i.e. in an Arithmetic Progression the numbers increase or decrease by a constant difference.
Each of the following series forms an Arithmetical Progression:
2, 6, 10, 14…
10, 7, 4, 1, -2…
a, a + d, a + 2d, a + 3d…

Example:
1. If the 7th term of an Arithmetical Progression is 23 and 12th term is 38 find the first term and the common difference.
Answer: 7th term = 23 = a + 6d —- (1)
12th term = 38 = a + 11d —- (2)
Solving (1) and (2) we get a = 5 and d = 3
2. How many numbers of the series 9, 6, 3 … should we take so that their sum is equal to 66?
Answer: n[18 + (n  1)3]/ 2 = 66
n2 – 7n – 44 = 0
–> n = 11 or -4.
The series is -9, -6, -3, 0, 3, 6, 9, 12, 15, 18, 21..
We can see that the sum of first 7 terms is 0. The sum of next four terms after 7th terms gives us the sum. Otherwise, if we count 4 terms backward from -9 we’ll get the sum as -66.
3. What is the value of k such that k + 1, 3k – 1, 4k + 1 are in AP?
Answer: If the terms are in AP the difference between two consecutive terms will be the same. Hence,
(3k – 1) – (k + 1) = (4k + 1) – (3k – 1)
2k – 2 = k + 2  k = 4.

TO INSERT ARITHMETIC MEANS BETWEEN TWO NUMBERS
Let n arithmetic means m1, m2, m3… mn be inserted between two numbers a and b. Therefore, a, m1, m2, m3, … mn, b are in arithmetic progression.
Let d be the common difference.
Since b is the (n + 2)th term in the progression, b = a + (n + 1)d
Whence d = (b – a)/(n + 1)
Hence m1 = a + (b – a)/(n + 1), m2 = a + 2(b – a)/(n + 1).. and so on.
Example:
4. If 10 arithmetic means are inserted between 4 and 37, find their sum.
First Method:
Let the means be m1, m2, m3… m10. Therefore 4, m1, m2, m3… m10, 37 are in AP and 37 is the 12th term in the arithmetic progression. Hence, 37 = 4 + 11d  d = 3
Therefore means are 7, 10, 13 … 34 and their sum is 205.
Second Method:
We know that in an AP-
the sum of first term + last term = sum of second term + second last term = the sum of third term + third last term = .. and so on.
Therefore, 4 + 37 = m1 + m10 = m2 + m9 = m3 + m8 = m4 + m7 = m5 + m6 = 41.
Hence m1 + m2 + m3… + m9 + m10= (m1 + m10) + (m2 + m9)…+ (m5 + m6)
= 5 x 41 = 205.

Example:
5. The sum of three numbers in A.P. is 30, and the sum of their squares is 318. Find the numbers.
Answer: Let the numbers be a – d, a, a + d
Hence a – d + a + a + d = 30 or a = 10
The numbers are 10 – d, 10, 10 + d
Therefore, (10 – d)2 + 102 + (10 + d)2 = 318
Or d = 3, therefore the numbers are 7, 10, and 13.

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Linear Equations
I am sure that this chapter is considered easy by many CAT 2007/ 2008 aspirants. But it is one of the most frequently occurring chapters in CAT. Mastery with equations helps you solve many problems in other areas also. As we progress deeper into Algebra on TG, we think it would be better to stat with this small but useful lesson. As your D- day comes nearer, revising some old concepts wouldn’t hurt you.
Any equation of the form p = 0, where p is a polynomial of degree one, is known as a linear equation. The polynomial p can contain a single variable, two variables, or more than two variables. Some examples of linear equations are given below:

Quant Challenge for CAT 2008 Aspirants
by pradeep pandey – Thursday, 21 February 2008, 09:58 PM

The following quant challenge has been prepared for all CAT 2008 aspirants by Mr. Pradeep Pandey, an experienced quant trainer of MBA aspirants and author of the book “Quantitative Aptitude for CAT.” Mr. P. Pandey has been creator of many interesting mathematics and data interpretation problems and he has graciously agreed to share many of them with students on TG to help them in their CAT preparation. So try these problems and post your solutions in this thread itself. Mr. P. Pandey will post the solutions to them later- Cat

Do you really want to crack CAT? Try these out first

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Quant Shortcuts for CAT 2007
I have written on this theme before also. But before CAT 2007 happens, I would like to remind all the TGites about these methods once again. What is a shortcut? By shortcuts, I do not mean a sudden inspiration or am erratic method which can be applied to only one or two problems. By shortcut I mean a set method, a pattern which can be repeated again and again. These shortcuts can be learned and applied under pressure also. The reason to apply these shortcuts is one- getting to answers fast and in an easier way than by actually solving the problems. I have discussed two basic methods here which I predominantly use:
1. Examine the options
2. Start small

Quant Shortcuts for CAT 2007
by Cat – Sunday, 11 November 2007, 04:23 PM

I have written on this theme before also. But before CAT 2007 happens, I would like to remind all the TGites about these methods once again. What is a shortcut? By shortcuts, I do not mean a sudden inspiration or am erratic method which can be applied to only one or two problems. By shortcut I mean a set method, a pattern which can be repeated again and again. These shortcuts can be learned and applied under pressure also. The reason to apply these shortcuts is one- getting to answers fast and in an easier way than by actually solving the problems. I have discussed two basic methods here which I predominantly use:
1. Examine the options
2. Start small

Learning Quant through problem solving
by Cat – Monday, 27 April 2009, 04:37 AM
Year after year, the only burning question which nags my mind is “how do I prepare students better for CAT quant?” Many of the answers have started coming. For example, after two years of TathaGat, we have realized that making our classroom teaching heavy with quality questions yields better results than having mere practice questions in the classroom content and having tests and mocks full of quality material. For one thing, many students don’t solve the mocks completely. For another, by the time mocks have arrived, they don’t have enough time to deliberate and develop the fundamentals behind good problems. Second thing which we have seen is that exposing students to a vast variety of questions on different fundas is better than giving repeated questions on the same funda. Although collecting problems of different flavors and colours is not easy, it is better than having repeated problems. Still, we have not worked everything out. There are questions which are unanswered. And all of them boil down to the same problem- how to make quant more easily understandable to students? Is quant really difficult to master for students who did not pursue it after school?
My vehement answer is no. It is not difficult to master. It takes only as much pain as it takes to master any other subject. And like any other subject, it becomes more and more understandable as you spend more and more time on it. But learning does not happen in classrooms. It happens when you look for new problems, try and discover new things, think deeply about some topics, experiment on your own, test your own theories, and so many other things. The key is to get totally involved. In this article, maybe my CAT aspirants will get some feel about how an aptitude for quant is slowly developed through different adventures.
I was recently having a look at a problem that I had posted in the Quant-DI forum long time back. The problem can be found in TG Special- 1 thread, a collection of my problems that I keep posting in the forum. The problem, shown in the figure below is extremely short:

Mensuration- Solids

For nearly every CAT 2008 aspirant, the question of ‘what to do and how to do it’ is the most crucial question of all. Besieged by too many topics and pulled in too many directions, a student is dazed and intimidated by the seemingly mammoth task at hand. The trick to winning a long-drawn-out battle like CAT preparation is to get started first and think later. Catch any end of the rope you can lay your hands on and start pulling. Some helpful and cultivated good habits, such as making a timetable for the next day and practicing all the three sections everyday, will also go a long way in preparing you for the D-day. And if you are one of them who forget what they read a month ago, you can certainly move around in TG forums and keep solving problems all the time to keep you in shape! But the most important thing to do is to find that zeal and excitement for your preparations in place of that CAT fear.
Needless to say, CAT preparations will bring you one of the most memorable times of your life. Even with all that stress, those class exercises, those depressing mocks and the fierce competition, CAT preparation is so much fun.

Given below are some teaser questions for you to try your hands on. Good luck!

Reply Greatest Integer Function and its Applications
by Total Gadha – Monday, 26 February 2007, 03:27 AM

The greatest integer function, denoted by [x], gives the greatest integer less than or equal to the given number x.
To put it simply, if the given number is an integer, then the greatest integer gives the number itself, otherwise it gives the first integer towards the left of the number of x on the number line.
For example,
[1.4] = 1
[4]= 4
[3.4] = 3
[ – 2.3] = – 3
[ – 5.6] = – 6, and so on.
NOTE: We can see that [1.4] = 1 + 0.4 or x = [x] + {x}, where {x} is the fractional part of x. For x = – 2.3, [x] = – 3 and {x} = 0.7
In the following figure, we can see that the greatest integer function gives the number itself (when the given number is an integer) or the first integer to the left of the number on the number line.

The graph of greatest integer function is given below. Note that the red dot indicates that integer value on the number line is not included while the green dot indicates that the integer value is included.

number system


Types of Numbers

Natural Numbers:
The group of numbers starting from 1 and including 1, 2, 3, 4, 5, and so on. Zero, negative numbers, and decimals are not included this group.
EXAMPLE
1. If n is an odd natural number, what is the highest number that always divides n(n2 – 1)?
Answer: n∙(n2 – 1) = (n – 1)∙n∙(n + 1), which is a product of three consecutive numbers. Since n is odd, the numbers (n – 1) and (n + 1) are both even. One of these numbers will be a multiple of 2 and the other a multiple of 4 as they are two consecutive even numbers. Hence, their product is a multiple of 8. Since one out of every three consecutive numbers is a multiple of 3, one of the three numbers will be a multiple of three. Hence, the product of three numbers will be a multiple of 8 ´ 3 = 24.
Hence, the highest number that always divides n∙(n2 – 1) is 24.
2. For every natural number n, the highest number that n∙(n2 – 1)∙(5n + 2) is always divisible by is
(a) 6 (b) 24 (c) 36 (d) 48
Answer:
Case 1: If n is odd, n∙(n2 – 1) is divisible by 24 as proved in the earlier question.
Case 2: If n is even, both (n – 1) and (n + 1) are odd. Since product of three consecutive natural numbers is always a multiple of 3 and n is even, the product n∙(n2 – 1) is divisible by 6. Since n is even 5n is even. If n is a multiple of 2, 5n is a multiple of 2 and hence 5n + 2 is a multiple of 4. If n is a multiple of 4, 5n + 2 is a multiple of 2. Hence, the product n∙(5n + 2) is a multiple of 8.
Hence, the product n∙(n2 – 1)∙(5n + 2) is a multiple of 24.
Hence, [b]
Rule: The product of n consecutive natural numbers is divisible by n!, where n! = 1 × 2 × 3 × 4 × 5…. × n
EXAMPLE
3. Prove that (2n)! is divisible by (n!)2.
Answer: (2n)! = 1•2•3•4• … •(n – 1)•n•(n + 1)• …•2n
= (n)!•(n + 1)•(n + 2)• …•2n.
Since (n + 1)•(n + 2)• …•2n is a product of n consecutive numbers, it is divisible by n!. Hence, the product (n)!•(n + 1)•(n + 2)• …•2n is divisible by n!•n! = (n!)2.
Whole Numbers:
All Natural Numbers plus the number 0 are called as Whole Numbers.
Integers: All Whole Numbers and their negatives are included in this group.
Rational Numbers: Any number that can be expressed as a ratio of two integers is called a rational number.
This group contains decimal that either do not exist (as in 6 which is 6/1), or terminate (as in 3.4 which is 34/10), or repeat with a pattern (as in 2.333… which is 7/3).
Irrational Numbers: Any number that can not be expressed as the ratio of two integers is called an irrational number (imaginary or complex numbers are not included in irrational numbers).
These numbers have decimals that never terminate and never repeat with a pattern.
Examples include pi, e, and √2. 2 + √3, 5 – √2 etc. are also irrational quantities called Surds.
EXAMPLE

Real Numbers: This group is made up of all the Rational and Irrational Numbers. The ordinary number line encountered when studying algebra holds real numbers.
Imaginary Numbers: These numbers are formed by the imaginary number i (i = √-1). Any real number times i is an imaginary number.
Examples include i, 3i, −9.3i, and (pi)i. Now i2 = −1, i3 = i2 × i = −i, i4 = 1.
EXAMPLE

Complex Numbers: A Complex Numbers is a combination of a real number and an imaginary number in the form a + bi. a is called the real part and b is called the imaginary part.
Examples include 3 + 6i, 8 + (−5)i, (often written as 8 – 5i).

Prime Numbers: All the numbers that have only two divisors, 1 and the number itself, are called prime numbers. Hence, a prime number can only be written as the product of 1 and itself. The numbers 2, 3, 5, 7, 11…37, etc. are prime numbers.
Note: 1 is not a prime number.
EXAMPLE
8. If x2 – y2 = 101, find the value of x2 + y2, given that x and y are natural numbers.
Answer: x2 – y2 = (x + y)(x – y) = 101. But 101 is a prime number and cannot be written as product of two numbers unless one of the numbers is 1 and the other is 101 itself.
Hence, x + y = 101 and x – y = 1. -> x = 51, y = 50.
-> x2 + y2 = 512 + 502 = 5101.
9. What numbers have exactly three divisors?
Answer: The squares of prime numbers have exactly three divisors, i.e. 1, the prime number, and the square itself.

Odd and Even Numbers: All the numbers divisible by 2 are called even numbers whereas all the numbers not divisible by 2 are called odd numbers. 2, 4, 6, 8… etc. are even numbers and 1, 3, 5, 7.. etc. are odd numbers.

Alphametics- The Art of Numbers
As CAT 2008 preparation starts heating up, I am going to flag the race with an oddball topic: Alphametics. What are alphametics or cryptarithms as they are also known? Alphametics are number puzzles in basic arithmetic operations where digits are represented by alphabets. It is customary that different digits are represented by different alphabets. Why are alphametics relevant? Because they require a good knowledge of properties of numbers and skill in working with them. Solving alphametics helps you quickly identify solution to many problems which might otherwise require a good amount of time. Let me give an example:

Can you solve this problem without any hit or trial method?

I hope the readers did not find the first article difficult to digest. The readers are advised not to hurry themselves. They are to go through these thoughts with absolute peace of mind, for the mind shall reveal its true potential one day for a good cause. No more lectures. Let us get back to mathematics. Have you noticed how a small seed holds the key to a big tree? Here is a question to prove the point-

To know the answer to this question, you will have to pay homage to the simplest of all rules- the digit-sum rule.
What is Digit Sum?
Given a number N1, all the digits of N1 are added to obtain a number N 2 . All the digits of N2 are added to obtain a number N3, and so on, till we obtain a single digit number N. This single digit number N is called the digit sum of the original number N1.
Example: What is the digit sum of 123456789?
Answer: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45 –> 4 + 5 = 9. Hence, the digit sum of the number is 9.
Note: In finding the digit-Sum of a number we can ignore the digit 9 or the digits that add up to 9. For example, in finding the digit-sum of the number 246819, we can ignore the digits 2, 6, 1, and 9. Hence, the digit-sum of 246819 is = 4 + 8 = 12 = 1 + 2 = 3.
Digit-Sum Rule of Multiplication: The digit-sum of the product of two numbers is equal to the digit sum of the product of the digit sums of the two numbers!
Example: The product of 129 and 35 is 4515.
Digit sum of 129 = 3 and digit sum 35 = 8
Product of the digit sums = 3 × 8 = 24 –> Digit-sum = 6.
Digit-sum of 4515 is = 4 + 5 + 1 + 5 = 15 = 1 + 5 = 6.
Digit-sum of the product of the digit sums = digit sum of 24 = 6
–> Digit sum of the product (4515) = Digit-sum of the product of the digit sums (24) = 6
Applications of Digit-Sum
1. Rapid checking of calculations while multiplying numbers
Suppose a student is trying to find the product 316 × 234 × 356, and he obtains the number 26525064.
A quick check will show that the digit-sum of the product is 3. The digit-sums of the individual numbers (316, 234 and 356) are 1, 9, and 5. The digit-sum of the product of the digit sum is 1 × 9 × 5 = 45 = 4 + 5 = 9.
–> the digit-sum of the product of the digit-sums (9) is NOT equal to digit-sum of the 26525064 (3)
Hence, the answer obtained by multiplication is not correct.
Note: Although the answer of multiplication will not be correct if the digit-sum of the product of the digit-sums is not equal to digit-sum of the product, but the reverse is not true i.e. the answer of multiplication may or may not be correct if the digit-sum of the product of the digit-sums is equal to digit-sum of the product
2. Finding the sum of the digits of a number raised to a power
Example: The digits of the number (4)24 are summed up continually till a single digit number is obtained. What is that number?
Answer: 43 = 64. Digit sum of 64 is = 1.
424 = 43 × 43 × 43 … × 43 (8 times)
Digit sums on both sides will be the same.
–> digit sum of 424 = digit sum of 1 × 1 × 1 × 1… (8 times) = 1

Example: Find the sum of the sum of the sum of the digits of 25!
25! = 1 × 2 × 3 × … × 24 × 25. As one of the multiplicands is 9, the digit sum will be 9.
3. Determining if a number is a perfect square or not

It can be seen from the table that the digit-sum of the numbers which are perfect squares will always be 1, 4, 9, or 7.
Note: A number will NOT be a perfect square if its digit-sum is NOT 1, 4, 7, or 9, but it may or may not be a perfect square if its digit-sum is 1, 4, 7, or 9.
Example: Is the number 323321 a perfect square?
Answer: the digit-sum of the number 323321 is 5. Hence, the number cannot be a perfect square.
Example: A 10-digit number N has among its digits one 1, two 2s, three 3s, and four 4s. Is N be a perfect square?
Answer: We can see that the digit sum of a perfect square is always 1, 4, 7, or 9. As the digit sum of the number is 3, it cannot be a perfect square.
Now can you answer the question that I posed? I bet that you can.

Suppose, the seed of any positive integer n is defined as follows:
Seed(n) = n, if n < 10
= seed(s(n)), otherwise,
Where s(n) indicates the sum of digits of n. For example,
Seed(7) = 7, seed(248) = seed(2 + 4 + 8) = seed(14) = seed(1 + 4) = seed(5) etc.
How many positive integers n, such that n y = 12 and hence x = 24

[Verfication: 10:00 PM + 11/3 hours + 24 * 11/3 minutes = 3:08 AM and similarly we can verify the time on the second watch also]

#2 – Rakesh Mahajan at 3:30 AM [a = 66/12 = 11/2, (x – y) = 8]
10:00 PM + 11/2 hours = 3:30 AM

So, this is possible only when both the watches lose time.

Total loss for Watch #1 = 3:30 AM – 3:08 AM = 22 minutes
Total loss for Watch #2 = 3:30 AM – 2:24 AM = 66 minutes

Therefore, x = 3y

Now, 3y – y = 8 => y = 4, and hence x = 12

[Verification: 10:00 PM + 11/2 hours – 11/2 * 4 = 3:08 AM. Similarly, we can verify the time on the second watch as well]

I think both these solutions satisfy all the conditions of the case let. Please let me know how can Chandraswami get away with the murder?

A milestone on a tortuous road..
It is funny how one problem gives rise to another. Of course this happens because of the irritating habit that mathematicians have of asking, “if it was something else and not something like this, what would be the solution?” And then you start all over again.
Here is a good problem I was discussing with a colleague-
In how many ways can you write the number 210 as a product of three integers?
210 = 2 × 3 × 5 × 7. Since the problem asks for integers, we will first calculate for positive numbers and then assign signs to these numbers.
Your first instinct is to write down 2, 3, 5 and 7 in a row and place two partitions between them. The partitions will give you three groups of numbers. Something like this:

But you should quickly realize that in this way no matter where you place your partitions, the number 2 and 7, 3 and 7, 2 and 5 will never be together. The partition method won’t work.
In fact, the situation is similar to placing 4 similar balls in three similar boxes. How would you do that? The hard way of course:
So we have to place 4 prime numbers in 3 places. If a place remains empty after distribution that means we will assume number 1 over there. Let’s find the number of ways of distributing these prime numbers. Since the box are all same only the different grouping of number matters.
As we are talking about integers, the numbers of possible cases are:
• All positive
• One positive and two negative (we will have to find different ways of assigning positive and negative signs also in this case)

There! 55 is the answer.
But do you stop at that? No, you ask another question:

So how do we do it?
The hard way again.
• Case I (positive, positive, positive)

• Case II (positive, negative, negative) e.g. (2, – 1, – 210)
The number of cases will be equal to the number of cases when all of them are positive. All we have to do is to write the integer triplets (x, y, z) of case I and then assign negative signs to y and z.
Therefore, total number of cases = 81.
• Case III (negative, positive, positive) e.g. ( – 2, 1, 210)

• Case IV (negative, negative, negative)
The number of ways in this case will again be equal to the number of ways in case III as we can assign negative sign to y and z to all the cases of case III. Therefore, the number of ways = 54
Therefore, the total number of ways = 81 + 81 + 54 + 54 = 270.
There. We have found answer to one more question. But what if it was…? The road is endless..

Some Mathematical Curios
by Total Gadha – Tuesday, 31 July 2007, 03:48 AM

Presented below are some of the findings that many CAT 2007 aspirants will find useful and many CAT 2008 aspirants will do well to add them in their armory early on. Unfortunately, I have not been able to provide the proofs for these results as the space, as well as the attention span of my readers, is at a premium. Perhaps, I will do so in my later chapters. Working as a teacher, I regularly come across useful results that can prove to be beneficial to one’s mathematical health but which do not reach the students because they are not properly documented. Well, hereon, I promise to document every useful result/proof that crosses the eyes and grey cells of this Gadha. Feel free to ask any question that arises in your curious mind.

Finding divisibility through seed numbers
Seed Numbers
Seed Numbers are used to find if a given number is divisible by a prime number. Although the concept is not used often, for some number, the divisibility rules cannot be applied and seed numbers come in handy there.
Every odd number (consider only odd prime numbers) gives unit digits of 1 and 9 in two of their first 10 multiples. For example, 3 × 3 = 9, and 3 × 7 = 21. For 17, 17 × 3 = 51, and 17 × 7 = 119. You can do this for any odd prime number and see that it is true. Now numbers ending in 1 or 9 can be written as multiples of 10 ± 1. For example 51 = 5 × 10 + 1, 119 = 12 × 10 – 1, etc. Now these numbers which are multipled by 10 (5 and 12 in this case) are the seed numbers for a particular prime number. The numbers are taken negative in the case of + 1 and positive in the case of -1. Therefore, every prime number has two seed numbers. In the above example, the seed numbers of 17 are -5 and 12. Here are seed numbers of some prime numbers listed down.
Numbers Multiples ending in 1 Multiples ending in 9 Seed numbers
3 21 = 2 × 10 + 1 9 = 1 × 10 – 1 – 2, 1
7 21 = 2 × 10 + 1 49 = 5 × 10 – 1 – 2, 5
13 91 = 9 × 10 + 1 39 = 4 × 10 – 1 – 9, 4
17 51 = 5 × 10 + 1 119 = 12 × 10 – 1 – 5, 12
19 171 = 17 × 10 + 1 19 = 2 × 10 – 1 – 17, 2
23 161 = 16 × 10 + 1 69 = 7 × 10 – 1 – 16, 7
How to use seed numbers:
Suppose you want to find out if 9044 is divisible by 17. You know that the seed number of 17 is -5. Here is what you do:
• Take out the unit digit of the number, multiply it with the seed number and add it to the number left after removing the unit digit. Therefore, take out the unit digit of 9044 i.e. 4, multiply it by -5, 4 × -5 = -20, and add it to the number left, i.e. 904. 904 – 20 = 884.
• Keep repeating this operation. For 884, take out 4, multiply it by -5 –> 4 × -5 = -20, add it to number left –> 88 – 20 = 68.
• In the end you will come up with a single digit or two digit number. If this number is divisible by the prime number, the original number is divisible by the prime number. Here, 68 is divisible by 17, therefore 9044 is divisible by 17.
Let’s do it once more.
Find if 43985 is divisible by 19.
We use the seed number 2 for 19.
• First step: 4398 + 5 × 2 = 4408
• Second step: 440 + 8 × 2 = 456
• Third step: 45 + 6 × 2 = 57
Now 57 is divisible by 19. Therefore, 43985 is divisible by 19.
You can use seed numbers to find divisibility by any prime number.
Divisors of a Number
How do we find divisors? For example, how do we calculate the number of divisors of 900?
Answer: 900 = 22 × 32 × 52. Therefore, any number that is a factor of 900 can have powers of 2 equal to 20, 21 or 22. Similarly for 3, the powers can be 30, 31, or 32 and for 5 they will be 50, 51, or 52. Writing the powers in a line we have-

Now any combination of a power of 2, a power of 3, and a power of 5 will give us a divisor. For example, in the figure, 21 × 32 × 51 will be a divisor of 900. As we can select a power of 2 in 3 ways, a power of 3 in 3 ways, and a power of 5 in 3 ways, the total number of combinations will be 3 × 3 × 3 = 27. Therefore, the number of divisors of 900 is 27.
Find the number of divisors of 15000.
Answer: 15000 = 23 × 3 × 54. The powers of various prime factors in a divisor can be:
2 – 20, 21, 22, or 23.
3 – 30 or 31
5 – 50, 51, 52, 53, or 54.

Again, the total number of combinations- and hence the total number of divisors- will be 4 × 2 × 5 = 40.
Find the number of odd divisors (divisors which are odd numbers) of 15000.
Answer: Odd numbers are not divisible by 2, and therefore in an odd divisor, there will not be a power of 2. Therefore, we will only consider powers of 3 and 5. Hence, the total number of combinations- and hence the total number of divisors- in this case will be 2 × 5 = 10
Find the number of even divisors of 15000.
Number of even divisors = Total number of divisors – number of odd divisors
= 40 – 10 = 30.
Find the number of divisors of 15000 that are perfect squares.
Answer: In a perfect square, power of every prime factor is even. Therefore, we will only consider those divisors of 15000 in which powers of prime factors are even. In other words, we will only consider even powers of prime factors. We write down only the even powers of prime factors

We can see that now the number of possible combinations = 2 × 1 × 3 = 6.
Therefore, 6 divisors of 15000 are perfect squares.
Find the number of divisors of 7! that are odd.
Answer: This question is same as the earlier question except that now we will first have to do prime factorization of 7!. For that, we will have to find powers of the prime factors in 7!. The prime factors present in 7! will be 2, 3, 5 and 7. Their respective powers are:

Therefore, 7! = 24 × 32 × 5 × 7. To find odd divisors of 7!, we ignore the powers of 2 and then calculate the number of combinations of powers of 3, 5, and 7. The number of combinations = 3 × 2 × 2 = 12. Therefore, there are 12 odd divisors of 7!.

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How to find the Units Digit

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How to find last two digits of a number

This brilliant article was contributed to Cat by Ashish Tyagi, a regular TGite. I do not think there can be easier or more lucid method of finding the last two digits of a number. Kudos to Ashish for this article. We will also welcome similar contributions from our other users. Do you have some useful or life-saving funda that you would like to share? Send it to us and we will publish it with your name- Cat

I am dividing this method into four parts and we will discuss each part one by one:
a. Last two digits of numbers which end in one
b. Last two digits of numbers which end in 3, 7 and 9
c. Last two digits of numbers which end in 2
d. Last two digits of numbers which end in 4, 6 and 8
Before we start, let me mention binomial theorem in brief as we will need it for our calculations.

Last two digits of numbers ending in 1
Let’s start with an example.
What are the last two digits of 31786?
Solution: 31786 = (30 + 1)786 = 786C0 × 1786 + 786C1 × 1785 × (30) + 786C2 × 1784 × 302 + …, Note that all the terms after the second term will end in two or more zeroes. The first two terms are 786C0 × 1786 and 786C1 × 1785 × (30). Now, the second term will end with one zero and the tens digit of the second term will be the product of 786 and 3 i.e. 8. Therefore, the last two digits of the second term will be 80. The last digit of the first term is 1. So the last two digits of 31786 are 81.
Now, here is the shortcut:

Here are some more examples:
Find the last two digits of 412789
In no time at all you can calculate the answer to be 61 (4 × 9 = 36). Therefore, 6 will be the tens digit and one will be the units digit)
Find the last two digits of 7156747
Last two digits will be 91 (7 × 7 gives 9 and 1 as units digit)
Now try to get the answer to this question within 10 s:
Find the last two digits of 51456 × 61567
The last two digits of 51456 will be 01 and the last two digits of 61567 will be 21. Therefore, the last two digits of 51456 × 61567 will be the last two digits of 01 ×21 = 21
Last two digits of numbers ending in 3, 7 or 9
Find the last two digits of 19266.
19266 = (192)133. Now, 192 ends in 61 (192 = 361) therefore, we need to find the last two digits of (61)133.
Once the number is ending in 1 we can straight away get the last two digits with the help of the previous method. The last two digits are 81 (6 ×3 = 18, so the tens digit will be 8 and last digit will be 1)
Find the last two digits of 33288.
33288 = (334)72. Now 334 ends in 21 (334 = 332 ×332 = 1089 × 1089 = xxxxx21) therefore, we need to find the last two digits of 2172. By the previous method, the last two digits of 2172 = 41 (tens digit = 2 × 2 = 4, unit digit = 1)
So here’s the rule for finding the last two digits of numbers ending in 3, 7 and 9:

Now try the method with a number ending in 7:
Find the last two digits of 87474.
87474 = 87472 ×872 = (874)118 ×872 = (69 × 69)118 × 69 (The last two digits of 872 are 69) = 61118 × 69 = 81 × 69 = 89
If you understood the method then try your hands on these questions:
Find the last two digits of:
1. 27456
2. 7983
3. 583512

Last two digits of numbers ending in 2, 4, 6 or 8
There is only one even two-digit number which always ends in itself (last two digits) – 76 i.e. 76 raised to any power gives the last two digits as 76. Therefore, our purpose is to get 76 as last two digits for even numbers. We know that 242 ends in 76 and 210 ends in 24. Also, 24 raised to an even power always ends with 76 and 24 raised to an odd power always ends with 24. Therefore, 2434 will end in 76 and 2453 will end in 24.
Let’s apply this funda:
Find the last two digits of 2543.
2543 = (210)54 ×23 = (24)54 (24 raised to an even power) ×23 = 76 × 8 = 08
(NOTE: Here if you need to multiply 76 with 2n, then you can straightaway write the last two digits of 2n because when 76 is multiplied with 2n the last two digits remain the same as the last two digits of 2n. Therefore, the last two digits of 76 × 27 will be the last two digits of 27 = 28)
Same method we can use for any number which is of the form 2n. Here is an example:
Find the last two digits of 64236.
64236 = (26)236 = 21416 = (210)141 × 26 = 24141 (24 raised to odd power) × 64 = 24 × 64 = 36
Now those numbers which are not in the form of 2n can be broken down into the form 2n  odd number. We can find the last two digits of both the parts separately.
Here are some examples:
Find the last two digits of 62586.
62586 = (2 × 31)586 = 2586 × 3586 = (210)58 × 26 × 31586 = 76 × 64 × 81 = 84
Find the last two digits of 54380.
54380 = (2 × 33)380 = 2380 × 31140 = (210)38 × (34)285 = 76 × 81285 = 76 × 01 = 76.
Find the last two digits of 56283.
56283 = (23 × 7)283 = 2849 × 7283 = (210)84 × 29 × (74)70 × 73 = 76 × 12 × (01)70 ×43 = 16
Find the last two digits of 78379.
78379 = (2 × 39)379 = 2379 × 39379 = (210)37 × 29 × (392)189 × 39 = 24 × 12 × 81 × 39 = 92
Now try to find the last two digits of
1. 34576
2. 28287

Modulo Arithmetic – Remainder Theory

This extremely useful article for finding remainders has been contributed to TG by Fundoo Bond, another intelligent TGite. All the CAT 2007/2008 who find this article useful are requested to say thanks to Fundoo Bond for his efforts and this wonderful compilation. If you have some interesting article, funda or helpful information to share, please mail it to us and we will post it by your name- Cat

The problems of finding the remainder are considered the most dreadful among the number theory problems. Fortunately, if we arm ourselves with some basic theorems, we will see that we can turn these mind-boggling problems into sitters and can ensure some easy marks.
I will be trying to share both the theory and the examples in order to make the concepts clearer. Again friends, do keep in mind that I am not a genius to write these theorems on my own. I have learnt them from various sources and I am trying my best to explain you in my own words.

Number System- 2: Remainders

Re: Number System- 2: Remainders
by Nikhil Rajagopalan – Thursday, 29 January 2009, 10:33 AM
Hi TG,
Thank you for this is excellent article for starters.
Regards,
Nikhil

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Re: Number System- 2: Remainders
by Arpita Bhattacharya – Sunday, 1 February 2009, 05:04 PM

if R=[(0.15)55+(0.25)55]/[(0.15)54+(0.25)54]
1.R>0.20
2.R>0.1
3.R=4
4.R0.20

[3^101]/77 find the reminder?

[17^36+19^36]/111 what is the reminder?

Got the ans of 3^101 /77 .
ans is 47.

1 and 8 are 2 natural numbers for which 1+2+3+….+n is a perfect square.

@venkat 49 and 288.
n.(n+1)/2 = x^2
=> n/2 and (n +1) both are perfect squares. So look for an odd square and half of its next number should also be a square
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Re: Number System- 2: Remainders
by nishant saxena – Saturday, 30 May 2009, 07:59 PM

hi sir
i wann aknow in ques 16 why u hav added 80 instead of subtracting it?????

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Re: Number System- 2: Remainders
by abhishek bindal – Sunday, 31 May 2009, 11:21 PM

@ hardik.. is the ans for your ques is 52.. do let me knoe
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Re: Number System- 2: Remainders
by Deepika Khandelwal – Tuesday, 2 June 2009, 10:05 AM

hi varun,
how to check :–
AB36 (A & B is any no. ) is perfect square this type of questions
please provide solution
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Re: Number System- 2: Remainders
by Deepika Khandelwal – Tuesday, 2 June 2009, 12:42 PM

hardik..how u got answer….. Got the ans of 3^101 /77 .
ans is 47.
could you please explain
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Re: Number System- 2: Remainders
by Priyesh Tungare – Tuesday, 2 June 2009, 05:25 PM

Hi,
In the question:
5. What the remainder when 2 96 is divided by 96?
y the common factor between 296 and 96 was 32? Can anybody please tell me y this two number were taken?

Also in the question where we have to find the remainder when 22004 is divided by 7, y did we took 8668? Please help me in this matter.

Thanks,
Priyesh Tungare
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Re: Number System- 2: Remainders
by Deepika Khandelwal – Wednesday, 3 June 2009, 11:58 AM

if we see that 96 is also even so definetly there will be common something
so 296 =25 291 and 96 is divisible by 32 so we have taken 32
as a common factor
I think I am able to solve your doubt

Hi
Cn you plz tel hw did u gt 47. ans shud be 26. please explain.

sorry guys i’v got 47.i was doing it in a diff. way

3^101 /77can be written as
(3^100 * 3)/77 = (81^25 *3) /77 =((77 +4)^25 * 3)/77
it can be now written as
(4^25 * 3)/77 = (4^24 *4 *3)/77
= ((77-13)^8 *4 *3)/77
= ((-13)^8 * 4*3)/77
=(13^8 *4 *3)/77
=((154 +15)^4 *4 *3)/77
=(15^4 *4 *3)/77
=((231 – 6)^2 *4 *3)/77
=(6^2 * 4 *3)/77
thus remainder comes 47
I hope I have not done error.Anybody with less lengthy process please reply.
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Re: Number System- 2: Remainders
by dindo saha – Friday, 5 June 2009, 11:02 PM

better method
(444^444)/7=(3^444)/7=(9)^222/7=(2)^222/7=(8)^74/7=1
best of luck
(soln courtesy to my roomate at IIT kgp hostel abhisek)
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Re: Number System- 2: Remainders
by dindo saha – Friday, 5 June 2009, 11:15 PM

(200!)/(100!)^2 remainder?
200!=1*2*3*4………..*200/(1*2*3*4…100)*(1*2*…..100)=
101*102*..200/(100!)
now 101*102*……200 has 100 elements .since product of consecutive n terms divisible by n! thus remainder=0
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Re: Number System- 2: Remainders
by Target IIM – Sunday, 7 June 2009, 03:09 PM

Hi Deepika,

Your approach is right. 296=25×3

so prob reduces to finding remainder of 291/3.Now applying Euler’s theorem we get remainder as 2.

Now we multiply 2 with 25(common factor) and hence the final ans is 64.
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Let S = 1!+2!+3!+4!+…+49! = 2!+3!+4!+5!+(6!+1)+7!+…+49!
since terms from 7!+..are multiples of 7
also according to wilson’s theorem 6!+1 is divisible by 7
therefore S= 2!+3!+4!+5! + 7*(some positive integer) = 152 + 7* (integer)
remainder = 5

find the remainder for ( 5^99) / 66

thanks saloni..but what if the divisor is 17 instead of 7.??is there any shortcut??

Ø(66) = 66(1 – 1/2)(1 – 1/33) = 32.
=> 532 = 1 mod66.
=> 599 = 532×3 + 3 = (532)3 x 53 = 53 mod66 = 125 mod66 = 59 mod66.

2^147 /11
i think ….
2^147 = 2^2 * 2 ^145 = (2^5)^71*2^2
(32)^71 *4 /11 = (33-1)^71 * 4/11 = (-1)^71 * 4 /11 = 40
but 40 /11 = 7 is the final remainder… B-

111=3*37
17^36=1(mod 37)
19^36=1(mod 37)
17^2=1(mod 3) therefore , 17^36=1 (mod 3)
||ly 19^36=1(mod 3)
so, (17^36+19^36)=2(mod 37)
(17^36+19^36)=2(mod 3)
hence (17^36+19^36)=2(mod 111)
Remainder = 2

What is the remainder of the 2000^1000 when divided by 13

Powers of a Number Contained in a Factorial

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Absolute Value (Modulus)- Basics

For all CAT 2008 aspirants starting their preparations, an introduction to absolute value (modulus) will help them to strengthen their basics. The credit for this chapter goes to my CAT 2007 students. Teaching in a classroom is a learning experience. A teacher learns more about the topic from students than what he learns from the books. This chapter is dedicated to all my CAT 2007 students who gradually, through their inquisitiveness, forced me to find more about the subject than what I already knew.

To understand absolute value function, we study the function from two different points.

To understand absolute value function, we study the function from two different points.

Base system

Suppose you have a 1 000 L tank to be filled with water. The buckets that are available to you all have sizes that are powers of 3, i.e. 1, 3, 9, 27, 81, 243, and 729 L. Which buckets do you use to fill the tank in the minimum possible time?
You will certainly tell me that the first bucket you will use is of 729 L. That will leave 271 L of the tank still empty. The next few buckets you will use will 243 L, 27 L and 1 L. The use of buckets can be shown as below

We can say that
1 000 = 729 + 243 + 27 + 1
= 1 × 3 6 + 1 × 3 5 + 0 × 3 4 + 1 × 3 3 + 0 × 3 2 + 0 × 3 1 + 0 × 3 0 .
The number 1 000 has been written in increasing powers of 3. Therefore, 3 is known as the ‘base’ in which we are expressing 1 000.
For example, The number 7368 can be written as 8 + 6 × 10 + 3 × (10) 2 + 7 × (10) 3 .
The number 10 is called the ‘base’ in which this number was written.
Let a number abcde be written in base p, where a, b, c, d and e are single digits less than p. T he value of the number abcde = e + d × p + c × p 2 + b × p 3 + a × p 4
For example, i f the number 7368 is written in base 9,
The value of (7368) 9 = 8 + 6 × 9 + 3 × 9 2 + 7 × 9 3 = 5408 (this value is in base 10)
There are two kinds of operations associated with conversion of bases:
1.7A conversion from any base to base ten
The number (pqrstu) b is converted to base 10 by finding the value of the number. i.e. (pqrstu) b = u + tb + sb 2 + rb 3 + qb 4 + pb 5 .
Example
38. Convert (21344) 5 to base 10.
Answer: (21344) 5 = 4 + 4 × 5 + 3 × 25 + 1 × 125 + 2 × 625 = 1474
1.7b conversion from base 10 to any base
A number written in base 10 can be converted to any base ‘b’ by first dividing the number by ‘b’, and then successively dividing the quotients by ‘b’. The remainders, written in reverse order, give the equivalent number in base ‘b’.
Example
39. Write the number 25 in base 4.

Writing the remainders in reverse order the number 25 in base 10 is the number 121 in base 4.
1.7c Addition, subtraction and multiplication in bases:
Example
40. Add the numbers (4235) 7 and (2354) 7
Answers: The numbers are written as

The addition of 5 and 4 (at the units place) is 9, which being more than 7
would be written as 9 = 7 × 1 + 2. The Quotient is 1 and written is 2.
The Remainder is placed at the units place of the answer and the Quotient
gets carried over to the ten’s place. We obtain

At the tens place: 3 + 5 + 1 (carry) = 9
Similar procedure is to be followed when multiply numbers in the same base
Example
41. Multiply (43) 8 × (67) 8
Answer:
7 × 3 = 21 = 8 × 2 + 5
7 × 4 + 2 = 30 = 8 × 3 + 6
6 × 3 = 18 = 8 × 2 + 2
6 × 4 + 2 = 26 = 8 × 3 + 2

For subtraction the procedure is same for any ordinary subtraction in base 10 except for the fact that whenever we need to carry to the right we carry the value equal to the base.
EXAMPLE
42. Subtract 45026 from 51231 in base 7.
Answer:

In the units column since 1 is smaller than 6, we carry the value equal to the base from the number on the left. Since the base is 7 we carry 7. Now, 1 + 7 = 8
and 8 – 6 = 2. Hence we write 2 in the units column. We proceed the same way in the rest of the columns.
1.7D IMPORTANT RULES ABOUT BASES
Rule1. A number in base N when written in base 10 is divisible by N – 1 when the sum of the digits of the number in base N is divisible by N – 1.
EXAMPLE
43. The number 35A246772 is in base 9. This number when written in base 10 is divisible by 8. Find the value of digit A.
Answer: The number written in base 10 will be divisible by 8 when the sum of the digits in base 9 is divisible by 8.
Sum of digits = 3 + 5 + A + 2 + 4 + 6 + 7 + 7 + 2 = 36 + A. The sum will be divisible by 8 when A = 4.
Rule2. When the digits of a k-digits number written in base N are rearranged in any order to form a new k-digits number, the difference of the two numbers, when written in base 10, is divisible by N – 1.
EXAMPLE
44. A four-digit number N1 is written in base 13. A new four-digit number N2 is formed by rearranging the digits of N1 in any order. Then the difference N1 – N2 when calculated in base 10 is divisible by
(a) 9 (b) 10 (c) 12 (d) 13
Answer: c
Reply

Percentage and its applications
by Cat – Wednesday, 4 April 2007, 07:31 PM

Your competence with percentages and its application should form a very important part of your armory of CAT preparation or, for that matter, preparation for any other MBA exam. The concept of percentage will be applied in not only your quant section but also your data interpretation (DI) section. Therefore, master this important concept and make it your habit to calculate percentages mentally.

The word ‘percentage’ literally means ‘per hundred’ or ‘for every hundred.’ Therefore, whenever you calculate something as a part of 100, that part is numerically termed as percentage.
In other words, percentage is a ratio whose second term is equal to 100. For example, 1:4 can be written as 25: 100 or 25%, 3: 8 can be written as 37.5: 100 or 37.5%, 3: 2 can be written as 150: 100 or 150%, and so on.
IMPORTANT CONCEPTS ASSOCIATED WITH PERCENTAGE
Basic formula of percentage:

Percentage of:

Percentage increase/decrease:
Percentage increase/decrease when a quantity a increase/decreases to become another quantity b is

Percentage less than/greater than:
Have a look at the picture given below:

You can see that Johnny is taller than Vicky. What will your answer be if I ask
(a) By what percentage is Johnny taller than Vicky?
(b) By what percentage is Vicky shorter than Johnny?
Answer:

Therefore, to find the final quantity after a 20% increase, we can directly multiply the old quantity by a factor of 1.2 and get the new quantity. Similarly, for a 20% decrease, we can multiply the old quantity by 0.8 and get the new quantity. The factors to be multiplied for various percentage increase/decrease are given below:

The biggest advantage of using the factors is that for subsequent percentage increase/decrease, we just keep on multiplying the corresponding factors and get the final quantity.
Example:
1. The performance bonus of a salesman increases by 10% in the first year, by 20% in the second year, and by 30% in the third year. What is the overall percentage increase in his bonus in 3 years?
Answers: Let the bonus at the start of the first year be Rs100.
Therefore, to find the final bonus we just multiply by factors.
Final bonus = 100  1.1  1.2  1.3 = 171.6.
Therefore, overall percentage increase = 71.6%

NOTE: note that taking initial value of 100 makes the problem simpler; whatever increase we get is directly equal to the percentage increase.
2. An amount was first reduced by 10% and then further reduced by 20% and Rs10 800 were left. What was the original amount?
Answer: Let the original amount be A.
Therefore A  0.9 (factor for 10% decrease)  0.8 (factor for 20% decrease) = 10 800
Or A = Rs15 000.
• SOME SOLVED EXAMPLES

Divisors of a Number
– Tuesday, 9 October 2007, 01:00 AM

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