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IIM Notification FAQ-6-August-2009

Summary:
Important Dates:
Advertisement for CAT 2009: Sunday, 30 August 2009.(in leading newspapers)
Sale of vouchers: 9 September 2009 to Thursday, 1 October 2009
(available at leading banks)
Online Registration: 9 September 2009 to Thursday, 1 October 2009
(Information available in the voucher)
CAT 2009 Test Dates: 28 November 2009 – Monday, 7 December 20009.
* You can choose your test date, session and test center
* Number of sessions: Two daily morning & afternoon
* No. of Test Centers: 30 cities

What does this mean for CIO users?
1. Make up your mind which session day of the week, session and test
center works best for you! We have always stressed that practice Mock
CATs in the exact exam like scenario. If you have been most
comfortable writing Mock CATs in 10 am - 12:30 pm time slots, then you
must reserve that slot for yourself. If you are most comfortable being
at home before CAT 2009, then choose test center closest to your home.
If Sundays work best for you as the test 'day', try to book that day
for your CAT 2009.

2. Once you have made up your mind regarding test date, center and
session, get yourself registered online ASAP. Any delay may rob you
off of your most optimal environment for test taking and may very
adversely impact your performance!!

We are committed to your best performance in CAT 2009! Because in your
spectacular success lies our success!

Non-IIM MBA Schools using CAT scores

There are lot of MBA institutes in India which takes CAT score as their criterion for admissions.So even if one doesn't get in to IIMs,still there is high chance to end up in some of prestigious MBA schools in India.Here is a list of Non-IIM MBA Schools using CAT scores

Concepts in Unit Digits for CAT QA and DI
================================
Unit digit concepts play a lot of time saving tricks in the CAT format.We can use the concept of unit digits in eliminating answer options in all the objective type competitive examinations.

The unit digit has got huge significance when it comes to eliminate the answer options.Lot of questions involving rigorous multiplications can be solved within no time by eliminating the answer options

CASE 1...
---------
Lets take the case of the product of two successive numbers,n(n+1)
We will see the possible unit digits of the products.

0*1=0
1*2=2
2*3=6
3*4=2
4*5=0
5*6=0
6*7=2
7*8=6
8*9=2

we can see that, the unit digits of products of successive numbers will end in 0,2,6 only.
Now we have the formula for sum of first n natural numbers, n(n+1)/2
So we will see the possible unit digits for these products.
We know that,n(n+1) ends in 0,2,6.
So we consider the possible digits, we will get, when we divides n(n+1) by 2.
0 gives 0,5
2 gives 1,6
6 gives 3,8
So these are the possible unit digits for the sum of first n natural numbers.

*Product of successive numbers,n(n+1) ends in 0,2,6

*The sum of first n, natural numbers will never end in 2,4,7,9

*Perfect squares never end in 2,3,7,8

*If there are n terms in an AP, the difference between sum of odd terms and sum of even terms is (n/2)*d where d is the common difference.

CAT Lessons-Progressions
======================

Progressions is a comparatively easy section that comes in CAT.
Once you understand the concepts in progressions, you can answer almost all the questions. Identifying that the questions belong to progressions is the most difficult part. Now we can go to the detailed study of the Progressions.
For a person aiming very high score in QA section,I would advise that he should definitely do the Progressions questions

Progressions can be divided into
------------------------------------------
Arithmetic Progression (AP)
Geometric Progression (GP)
Harmonic Progression. (HP)
Arithmetic Geometric Progression. (AGP)



Arithmetic Progression (AP)
-------------------------------------------

In Arithmetic Progression, the successive terms always has a constant difference between them. The example of an AP is given below.

1,3,5,7,9…….This is an AP series with first term (a or t1)=1 and the constant difference (common difference or cd)=2.
2,4,6,8,10…… This is an AP series with first term (a or t1)=2 and the constant difference (common difference or cd)=2.

The nth term for an AP with first term =‘a’ and cd=’d’ is
tn =a+(n-1)d.
The sum of first n terms of an AP is
Sn=n/2{2a+(n-1)d}


Take the case of AP,2,4,6,8,10……

The 5th term can be found out by
t5=2+ (5-1)2
=2+4*2
=2+8
=10.
As we can see from the series the 5th term is indeed 10.

The sum of first 5 terms can be found out by
S5=5/2{2*2+(5-1)*2)
=5/2{4+8}
=5*6
=30.

Also we can see that sum of first 5 terms are 2+4+6+8+10=30.

Arithmetic Mean (AM)
----------------------------------
AM of an AP with terms a and b is AM = (a+b)/2.
Between two terms of an AP we can put any number of AMs.

Eg.
Suppose we need to put 4 AMs between a and b.,
Now the total number of terms in AP is 6.
Now the common difference in the AP is (b-a)/(4+1)= (b-a)/5..


Properties useful in solving questions in APs
----------------------------------------------------------------------
****
If in an AP, mth term is n and nth term is m, then (m+n)th term is always zero.
Eg. Consider the AP 4,3,2,1,0,-1,-2…..
1st term is 4 and 4th term is 1,therefore (4+1) term ie 5th term ,according to the property should be 0.We can see that the property holds true here.

****
If in an AP, sum of first m terms is equal to sum of first m tems, then sum of first (m+n) term is always zero.
Eg. Consider the AP -2.-1,0,1,2,…..
Sum of first 1 term is -2.
Sum of first 4 terms is -2.
Therefore according to property, sum of first (1+4) ie 5 terms should be 0.
We can verify that the property holds true.

Geometric Progression (GP)
----------------------------------------

In Geometric Progression, the successive terms always has a constant ratio between them. The example of an GP is given below.

1,3,9,27,81…….This is a GP series with first term (a or t1)= 1 and the constant ratio (common ratio or cr)=3.
2,4,8,16,32…… This is a GP series with first term (a or t1)= 2 and the constant ratio (common ratio or r)=2.

The nth term for an GP with first term =‘a’ and common ratio=’r’ is
tn =ar^(n-1)
The sum of first n terms of a GP is
Sn=a{r^(n-1)/(r-1)} r>1
&
Sn = a{r^(n-1)/(1-r)} r<1

Geometric Mean (AM)
------------------------------
GM of a GP with terms a and b is GM = (a*b)^0.5
Between two terms of an GP we can put any number of GMs.

Eg.
Suppose we need to put 4 =GMs between a and b.
Now the total number of terms in =GP is 6.
Now the common ratio of the GP is (b/a)^(4+1)= (b/a)^0.5

But in Competitive exams like CAT,GMAT etc, in GP, sum to infinity is more significant than sum of n terms.
Sum to infinity of GP is infinity if r>1,so the cases where, r<1 only we will get a definite sum.
Sum to infinity of a GP with first term,a and common ratio,r is
S∞=a/(1-r).

Eg. Find the sum to infinity of the series 1,1/2,1/4,1/8,1/16…….
a=1,r=1/2
S∞=1/(1-1/2)=2.

Harmonic Progression
-----------------------------------

If a,b,c are in Harmonic Progression(HP),then 1/a,1/b,1/c are in AP.
This is the standard definition of an HP.


Also we have the general rules
1. AM>=GM>=HM
2. GM^2=AM*HM



Strategies for solving problems in Progressions.
----------------------------------------------------------------------

**
In all questions that has Tn and Sn variable, n, always put values for n and check for the answer option.

**
Always try to make up a progression satisfying the conditions given in the questions and solve that progression only instead of doing with unknown numbers.

**
Suppose we need to take one AP,one GP and one HP having 3 terms with same first and last terms,then we an take the following progressions,

AP: 1, 2/3,1/3 cd=1/3
GP : 1,1/(3^.5),1/3 cr=1/(3^.5)
HP: 1,1/2,1/3 HM=1/2.

**
Find the 625th term of the series 1,2,2,3,3,3,4,4,4,4,5………

Analysing the series, we see

Last 2 is the 3rd term ie 2*3/2
Last 3 is the 6th term ie 3*4/2
So we got the pattern,Last n will be the n(n+1)/2 th term
Also the previous n terms will be same,n only.

So if we consider,the case of 625th term, consider 35*36/2 gives 630.
Means 630th term will be 35,also previous 35 terms will be same 35 only.
So our answer is 625th term is 35.

Data Sufficiency Questions
=====================

Data sufficiency questions are a good source for scoring marks in competitive entrances using very less time.The advantage is that you never has to find the exact answers.You have to find whether the given statements are enough to find out the answer to the question.
This needs a bit of practice,since we need to analyse the given statements,individually first and then combined form.Also the answer options will be tricky sometimes.But practice it and you can increase your score easily .Also in lot of exams like CAT,GMAT,XAT etc,Data Sufficiency appears in both Quantitative section and Data Interpretation section.So its very easy to clear the cut-off marks if you can convert 4-5 DS questions.Also DS is now appearing in Bank entrances also
Here are some examples of DS questions for competitive exams...

Data Sufficiency Questions
----------------------------------

Answer the following questions by marking the answer options a,b,c,d or e.
a.The question can be answered by statement 1 alone.
b.The question can be answered by statement 2 alone.
c.The question can be answered by either statement alone.
d.The question can be answered only by combining both statements together.
e.The question cannot be answered by given data.


Who types at a faster rate, Raju or Renju?
(1) The difference between their typing rates is 10 words per minute.
(2) Renju types at a constant rate of 80 words per minute.

What was Anil’s average (arithmetic mean) grade for all of his courses?
(1) His grade in social studies was 75, and his grade in science was 75.
(2) His grade in mathematics was 95.

If today the price of an item is $3,600, what was the price of the item exactly 2 years ago?
(1) The price of the item increased by 10 per-cent per year during this 2-year period.
(2) Today the price of the item is 1.21 times its price exactly 2 years ago.

If the Eden ground is rectangular, what is its width?
(1) The ratio of its length to its width is 7 to 2.
(2) The perimeter of the playground is 396 meters

What is the value of x –1?
(1) x + 1 =3
(2) x – 1 <> r – s?
(1) x > r and y < y =" 2," s =" 3," r =" 5," x =" 6." y =" 7" y =" 3"> 9
If x and y are consecutive odd integers, what is the sum of x and y?
(1) The product of x and y is negative.
(2) One of the integers is equal to –1.

What is the value of x?
(1) 3 + x + y = 14 and 2x + y = 15
(2) 3x + 2y = 12 + 2y

If John is exactly 4 years older than Anil, how old is Aby?
(1) Exactly 9 years ago Aby was 5 times as old as Anil was then.
(2) Anil is more than 9 years old.

CAT Data Interpretation-Interesting Question
============================================
Matchstick problem
----------------------------------

The Matchstick problem is a cult problem, which is frequently appearing in TIME-AIMCATs and its variants appear in lot of competitive exams like CAT,XAT,MAT, Campus placement papers….This problem appears in various forms,like two people going for a drive taking turns while driving for a max. & min. kilometer limits.
The variants of same question came 2-3 times in original CAT papers.

Mainly 2 variants are there for this problem, we will discuss the type 1 variant in this post.

Type 1.
=================

There are n matchsticks, and 2 players A and B. One person should take minimum of 1 stick and can take maximum of 5 sticks at a time. The person who takes the last stick is the loser. Each player will play intelligently in order to win.
a.If there are 10 matchsticks and A has to play next, how many sticks he has to take to ensure that he wins?
b.If there are 21 matchsticks, and B is to play, Is there a chance for B to win?

Explanation:-

Consider the situation of 8 matchsticks and A has to play.
A can take max 5 and min 1.
Consider the sequence of steps that will follows if A plays intelligently to win

1. A will take 1 and remaining is 7.
2. Now B can take a max of 5, suppose he takes 5,then ,2 will remaining.
3. A will take 1 and 1 will remain.
4. B has to take min. 1.so he will lose and A will win.

Consider another sequence where in step 2,B takes min 1.

1. A will take 1 and remaining is 7.
2. Now B takes 1,then ,6 will remaining.
3. A will take 5 and 1 will remain.
4. B has to take min. 1.so he will lose and A will win.
So in both these extreme cases, irrespective of B’s play, A is winning.

This leads us the conclusion that, in this problem,
‘Winner is decided by initial number of matchsticks and the first play.”

Also, the aim of the player who plays first will always be to leave
{(min. limit + max. limit)+1}
ie here {(1+5)+1}=7 numbers of matchsticks to his opponent.
The same scenario will occur even if he leaves the multiples of this number.
ie in our question, the player who plays first, should aim to leave either 7 or a multiple of 7 matchsticks to his opponents so that irrespective of the future moves he is sure about his success.
So, in a game with the above min and max limits, the first player should aim at leaving the number of matchsticks as 7,14,21,28,35……..

The aim of the player who plays first will always be to leave
{(min. limit + max. limit) +1} numbers of matchsticks to his opponent.

Note:
Here if initially the number of matchsticks is a multiple of 7, then the first player will always lose.

Variant of Type 1:
----------------------------
In the above case the min. limit was 1.
Suppose the min. limit has been increased to a higher number,say2.
Then also we need to apply our general formula,
ie the aim of the player who plays first will always be to leave
{(min. limit + max. limit) +1} or {(min. limit + max. limit) +2} numbers of matchsticks to his opponent.

So in this case, the player who plays first, should aim to leave either 8 or 9 or a multiple of 8 or 9 matchsticks to his opponents so that irrespective of the future moves he is sure about his success

Type 2 --will be discussed in coming posts...

CAT Lessons--Quantitative Aptitude
================================

Targeting High marks in CAT,Improve your Quant.....

Numbers
=========
In last post we discussed about finding unit digit of a large power using Fermat's theorem and co-prime Technique.Now move 1 step forward,Find last 2 digits.
To find tens digit or last two digits, we will follow the same techniques except for this time we will divide by 100. 100 has 40 relative primes.
100(1-1/2)(1-1/5)=40

So to find tens digit of 69^83 we will divide 83 by 40 [power by relative primes]. We will get 3 as remainder. So we are left with 69^3 divided by 100.
this can be done manually by multiplying only the last 2 digits.
we will get 69.

Similarly we can do it for last three digits, method remains same but this time we will divide by 1000.
also remember,no. of co-primes for 1000=1000(1-1/2)(1-1/5)=400


Some examples are given
========================
* “What will be the remainder when 68^66 is divided by 5”.

First find relative primes of 5 [ 5 is prime so its relative primes are 4]
Now divide the power of 68^66 by 4, so we get 2.
Now we are left with 68^2 divided by 5.
Do 8^2/5, will get remainder as 4.

Suppose the base of numerator and denominator are not relative primes then take out the relative prime from Nr and Dr and multiply it in the end.


*Find the remainder when 78^67 is divided by 8.

Here we take out 2 from base of Nr and Dr. Hence we are left with
2*{39^67/4}.
Applying our usual method to get remainder as 3 multiplying with 2 to get the result as 6.


More posts on quant and DI are coming soon......

Online CAT coaching:-
====================
Quantitative Analysis :Numbers,Remainder questions,Fermat's Theorem
---------------------------------------------------------------------------------------


Fermats theorem:-

This theorem is very important for solving the remainder questions with large powers or with large divisor. This can be used easily in solving almost all CAT level difficult remainder questions quickly.

The Theorem is as given below.on

If a & b are co-primes or relative primes and b is prime then
rem{a^(b-1)/b}=1.

If the divisor ,b is composite, then find n, the no. of co-primes between 1 and the no. Now we get rem(a^n/b) is 1.

Two relative prime numbers are those which have no common factor besides 1, such as 57 and 59], to find relative primes of number N we can use the formula N(1-1/p1)(1-1/p3)(1-1/p2)...................(1-1/pn). Where p1,p2,pn are prime factors of N.

Relative primes of 12 are 12(1-1/2)(1-1/3)=4 and those are 1,5,7,11
here 2 and 3 are the prime factors of 12.

Some practice questions are given here..

Eg.1. Find the rem(3^40/13)?

Since 3 & 13 are co-primes also 13 is prime, using Fermats theorem,rem(3^12/13)=1.So rem(3^36/13) =1.Therefore now we need to find out rem(3^4/13).
ie rem(81/13)=3.

Eg. 2. Find the rem(5^72/11)?

Since 5 & 11 are co-primes also 11 is prime, using Fermats theorem,rem(5^10/11)=1.So rem(5^70/11) =1.Therefore now we need to find out rem(5^2/11).
ie rem(25/11)=3.

Eg.3. Find the rem(5^72/12)?

Here 12 is not a prime no.
12=2^2*3.
No. of co-primes of 12 is 12(1-1/2)(1-1/3)=4.
Therefore rem(5^4/12)=1.
So rem(5^72/12)=1.

Find the Unit digit of 69^77?


What we need to find is the rem{(69^77)/10},since it will be the unit digit.
Now no. of relative primes of 10 is 10(1-1/2)(1-1/5)=4

According to Fermat's theorem,rem(69^4)/10 will be 1
Now what we do is divide the power(77) with number of relative primes(4).
So divide 77 by 4 you will get 19 remainder 1. So you are left with 69^1 divided by 10 and that is easy. So the remainder is 9 [89/10 gives 9 as remainder].
And 9 is your unit digit.

OR

But,this question can be done very easily by another method,
Use power cycle.
unit digit 9 has power frequency 2(9,1)
9 raise to any odd number is 9 only.
so the remainder(here unit digit is 9 only).

Important Number Questions
===========================
Some typical number questions are given in this posts.The variants of the same questions are frequently coming in CAT and various other competitive exams like XAT,GMAT,MAT,IIFT,Symbiosis....So have a look at these questions...


1*Find the number of factors 24 has?

Explanation:-
24 can be written as 2^3*3.
We know that if N=a^m*b^n. Then the number of factors of N is (n+1)*(m+1)
Here we have, number of factors= (3+1) (1+1) = 8.

2*Find the least number with no. of factors as 24.

Explanation:-
A number can have 24 factors,
If its in the form, N=a^23,
or a*b^11
or a^2*b^7
or a^3*b^5
or a*b*c^5
or a*b*c*d^2 .

Here if we want to find the smallest number we need to fit in the smallest prime numbers as far as possible.
So better if you put the last choice, a*b*c*d^2, also put d=2,c=3,b=5,a=7
So the number becomes 2^2*3*5*7 ie
4*3*5*7= 420.
So this is the smallest number with 24 factors.
If we put any other forms of the number, the resulting number will be higher than 420.

*A number has unequal prime factors and the total number of factors is 4.
The sum of the factors without 1 and the number itself is 30. Find the smallest number with this characteristics.

Explanation:-
The number with 4 factors can be with the form a^3 or a*b.
Since its given that number has unequal prime factors, it can take the form of a*b.
Then the factors are 1, a, b, ab where ab is the number itself.
The condition given is a+b=30, where a and b are prime numbers.
a,b can be 7 & 23
11 & 19
13 & 17.
Then the smallest number is 7*23 = 161.

*How many numbers are there <1000, which has exactly 3 factors.

Only squares of prime numbers have exactly 3 factors.
So we need to find the number of prime number squares<1000.
The highest prime number square less than 1000 is 31^2.
So all prime number squares up to 31 can be regarded as the answer.
They are squares of 2,3,5,7,11,13,17,19,23,29,31.So there are total 11 numbers.
So answer is 11.

Feel free to post new questions in these topics..

Data Interpretation for CAT--Introduction & Basic Fundas
===================================================

Data interpretation is one most critical section in CAT and other management exams which can become a nightmare for a lot of students. In this section, different caselets with 4-5 questions will be there. The questions and pattern in this section may change year after year, so always make sure that you score the cut-off marks for DI section and be safe.

The questions can be normal puzzles, numbers based puzzles, pattern matching, graph based caselets, game based caselets….

You must be fast in calculations, especially with fractions and percentages, in order to score high marks in DI section. Also in lot of graph based questions, you can answer them by mere observation alone.

If you analyze the DI questions, in recent CATS, a lot of reasoning questions are also appearing in CAT DI.

Some important concepts helpful in CAT-DI

Percentage concepts
===================

When a number becomes double, we say that it increased by 100%.
Also if it becomes thrice,the number increased by 200%.
ie the percentage increase ,if number got multiplied by ‘n’ times is
(n-1)*100.
If the number increases 14 times,the percentage increase is (14-1)*100=1300%

Similarly,if we say that a number got percentage increase of 1200,we mean that it became,{(1200/100) +1} ie 13 times its original value.

Also finding the squares of numbers, quickly is another important need for scoring high marks in DI section.

Fast way of finding Squares of numbers.
===================================
For finding squares of all numbers,we have a rather easy way.

Eg 1.Find the square of 32.

(32-2)*(32+2)+2*2= 30*34+4=1020+4=1024.

The advantage of this method is that we are converting the squaring procedure to a single digit multiplication plus addition. This is quite easy compared to the usual 2-digit multiplication.
In the example given above,the 32*32 multiplication is changed to a simple single digit multiplication of 34*3. and addition.

Eg. 2. Find the square of 26.

(26-6)*(26+6)+6*6=20*32+36=640+36=676
or this can be done in one more way,
(26+4)(26-4)+4*4=30*22+16=660+16=676.

Both ways are easier than the original multiplication of 26*26.

We can do any square using this method; however we need to improve our speed in single digit multiplication.

Faster way of finding Square-roots of numbers
=========================================

Finding approximate roots of numbers can also be useful in DI sections.
We have an easy method for that too.

Eg1. Suppose we need to find the root of 30.

The largest number with its square within 30 is 5(5*5=25).
Now 30/5 is 6.
So square root of 30 is (5+6)/2 = 5.5

If you take the square root of 30,it will come to 5.477. Approximately 5.5.

Eg2. Find the root of 40.

The largest number with its square within 40 is 6(6*6=36).
Now 40/6 is 6.667
So square root of 40 is (6+6.667)/2 = 6.333.
If you take the square root of 40,it will come to 6.324. Approximately 6.333.

Quantitative Aptitude Techniques.
--------------------------------------------
Here we will discuss some more quantitative methods for typical MBA examination questions.


Highest power of a number in a factorial.
--------------------------------------------------

This is a typical CAT question, which is being asked for many competitive entrance exams.
To find out the highest power of a no. in a factorial, we should find the no. of times the particular no. is repeating in that factorial. Hence we will divide the factorial no. continuously until we get a no. less than the original divisor.


Eg.Find the highest power of 5 in 100!

5)100
5)20
5)4.
Therefore the max. power of 5 in 1001! is 20+4=24

If the divisor is a composite no, we need to find the highest power of all the prime no. components and we can fix the highest power of the no, as the power of largest prime num. component.

Eg. Find the highest power of 10 in 100!
As we know 10 =5*2( both are prime numbers.)
Also 5>2.
So we need to find the highest power of 5 in 100! And this will be the answer.
(Anyway the highest power of 2 is more than that of 5, but we can’t take that since,
for getting a 10 we need both 5 and 2, so whichever is least, only that many 10s will be there).

Number of zeroes in a factorial.(n!)

The num. of zeroes will depend on the highest power of 10 in that factorial. That in turn depends on the highest power of 5( as we mentioned earlier.)

****In any number system, the number of zeroes depends on the highest power of the base number of that factorial****




Expressing a number as the difference between 2 squares.
--------------------------------------------------------------------

Suppose we have a number N=a.b
We can express N as
N= a.b=(a+b/2)^2-(a-b/2)^2
So we conclude that
In order for N to be an integer, both a &b should be either odd or even.
****So any multiple of 4 can be expressed as the difference between 2 squares.****

Lets take an example.

a.How many ways we can express 24 as the difference between 2 squares?
Explanation:-

24=1*24(format N=a*b)
24=2*12
24=3*8
24=4*6

Here 2*12 and 4*6 are the 2 forms where both a and b are even or odd. So 24 can be expressed as the difference between 2 squares in 2 ways.

2*12 as 7^2-5^2
4*6 as 5^2-1^2

Odd
-----
All odd prime no. can be expressed in only 1 way
Odd composite no. can be expressed in more than 1 way

Even
-------
4 can be expressed in only 1 way.
Prime multiple of 4 can be expressed in only 1 form.
Other multiples of 4 can be expressed in more than 1 form.
Non multiples of 4 can’t be expressed as difference between 2 squares.

Eg.
How many no. are there below 1000,that can’t be expressed as a difference between two squares?
Explanation:-

All odd num. can be expressed as the difference between two squares.
In even num., only which are non-multiples of 4 can’t be expressed.
So in 999 num. 499 even numbers are there.
Out of which ,250 are multiples of 4.
So non- multiples are 499-250=249.
So the answer is 249.

Online CAT Lessons
-------------------------

In CAT ,in Quantitative section,we can save a lot of time by eliminating the answer options,and divisibility checks comes in handy in this process.especially if you know the checks for prime numbers like 11,13,17...
So lets check out this quantitative methods...

Divisibility Checks for 11

We know that rem (10^1/11)=-1
Therefore the rem(10^2/11)=+1

So for a no. in the form ‘abcdef’, the divisibility can be checked by taking the diff. b/w
Sum of the odd numbered digits and even numbered digits.Check whether that difference is divisible by 11.If that is divisible then the large no. is divisible by 11.

Eg. Take the no. 1331
D
Odd numbered dig. Are 1,3 from right, sum =4
Even numbered dig. are 3,1 from right , sum =4
Difference of the sum is 0. So its divisible by 11

Divisibility Checks for 13

We know that rem (10^3/13)=-1
Therefore the rem(10^6/13)=+1

So for a no. in the form ‘abcdef’, the divisibility can be checked by taking the diff. b/w
abc and def. Check whether that difference is divisible by 13.If that is divisible then the large no. is divisible by 13.

Eg. Check the divisibility of 214175?
Split the no. in to 2 as 214 and 175.
214-175= 39
Since the difference 39 is divisible by 13, the no. 214175 is also divisible by 13.

Divisibility Checks for 17

We know that rem (10^9/17)=-1
Therefore the rem(10^18/17)=+1

So for checking the divisibility of 17,we need to take blocks of 9 digits and do the same procedure as we do for that of 7 or 13.

The knowledge of divisibility of various numbers will help you to eliminate some answer options without doing a single calculation. This types of small time savings can make a big difference in the whole exam.(As I said earlier, reaching answer quickly is the important thing…..)

Online CAT Lessons:-
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Divisibility Checks for various digits.
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Its always advantageous to know the divisibility rules of various digits so that we can save a lot of time during our calculations and its extremely useful for elimination of answer choices.
In an exam like CAT or GMAT, the most important thing is not to arrive at the answer, but arrive at the answer fast if possible, without doing any calculation you will have to find the answer. So,when you get an objective type question,uy first aim should be to eliminate at least 2 options. The knowledge in rules like divisibility, factors etc will come handy there.

The basic rule in finding the divisibility of a number is that we should find a power of 10(if we r using decimal system) which is completely divisible by the specific digit.
Or we need to find the power which gives a rem. of +/-1.

Divisibility Checks for 2
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We know that rem (10/2)=0.
ie rem(10^1/2) = 0.So we need to check the last digit of a no. only for checking its divisibility by 2.
****The last digit should be 2,4,6,8 or 0.****

Divisibility Checks for 3
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We can write any no. in decimal abc in the form 100a+10b+c
Again, abc = 99a+9b+(a+b+c)

Here except the last term,all terms are divisibe by 9 hence also by 3
Hence if the term ( a+b+c) which is the digit sum of the no. is divisible by 3,the no.abc is divisible by 3.

****The digit sum of a no. should be a multiple of 3****

Divisibility Checks for 4
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We know that rem (10^2/4)=0.
So we need to check the last 2 digits of a no. for checking its divisibility by 4.

****The last 2 digits should be divisible by 4.****

Divisibility Checks for 5
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The last digit should be 5 or 0.

****The last digit should be divisible by 5 or 0.****


Divisibility Checks for 6
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6=2*3.
A no. is divisible by 6,if its divisible by both 2 &3.

****The no. should be divisible by both 2 &3****

Divisibility Checks for 7
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We know that rem (10^3/7)=-1.
therefore rem(10^6/7) = +1.
So if we need to check the divisibility of a 6 digit number 'abcdef'
we need to split the number in to 2, 'abc' and 'def'
Now we need to take the difference between the two numbers and check whether its divisible by 7.If its divisible,then the large number is divisible by 7.

Divisibility Checks for 8
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We know that rem (10^3/8)=0.
So we need to check the last 3 digits of a no. for checking its divisibility by 8.

****The last 3 digits should be divisible by 8.****

Divisibility Checks for 9
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We can write any no. in decimal abc in the form 100a+10b+c
Again, abc = 99a+9b+(a+b+c)

Here except the last term,all terms are divisibe by 9.
Hence if the term ( a+b+c) which is the digit sum of the no. is 9,the no.abc is divisible by 9.

****The digit sum of a no. should be 9****

The divisibility check for higher prime numbers will be given in next lesson

Lets start with the most important section in Quantitative section,Numbers..

Numbers
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Numbers has lot of properties to be studied in detail and certain set piece questions are there.from where CAT questions are frequently coming..

Power cycle
Power cycle for digits 2,3,4,5,6,7,8,9

Divisibility checks
2,3,4,5,6,7,8,9,11,13…

Remainder questions
a) Using power cycle
b) Using binomial theorem

Highest power of a number in n!

Problems related to factors of a number
No. of factors.
Sum of factors
No. of co-primes
Product of factors

Express a no. in different forms
Expressing as difference of squares.

Fermats theorem
Eulers no.
Solving remainder questions using Fermat’s Theorem

Miscellaneous properties
1.Properties of a^n- b^n when ‘n’ is odd,even…
2.n^p-p when p is prime.

a. Lets start with Power cycles

****The frequency of the power cycles of various digits is very important.
This helps you to solve remainder questions involving large powers.*****

Let us take some example.

1. Find the remainder of 3^75/5?

Explanation:-

We know the power cycle frequency of 3.Its 4and the power cycle is 3,9,7,1
So convert the large power in the question (here 75) to a much smaller number using this freq. 75=4*18+3
Therefore 3^75 is now 3^{(4*18)+3},this can still be shortened as 3^3 only.
Since the other power frequencies will be repeating only.
Hence our question reduces to finding rem. of 3^3/5
This is quite easy, 27/5,rem. is 2.(or since we know power cycle of 3,3^3 ends with 7,so remainder is 7/5=2).

2.Find the remainder of 2^102/3?

Explanation:-
We know the power cycle frequency of 2.Its 4and the power cycle is 2,4,8,6.
Now we convert the large power in the question (here 102) to a much smaller number using this freq. 102=4*25+2.
Therefore 2^102 is now 2^{(4*25)+2},this can still be shortened as 2^2 only.
Since the other power frequencies will be repeating only.
Hence our question reduces to finding rem. of 2^2/3.Its 4 /3,remainder is 1.

OR

This question can be done in a very easy method.
We knew that rem. of 2/3 is -1.

3.Find the rem. of {(2^203)*(3^506)}/5?

Explanation:-
Here we can use the fact that ,

*****The rem. of a product is the product of the individual rem.****

So the ques. changes to rem(2^203)/5 * rem.(3^506)/5
Rem(2^203)/5 can be found out by using power cycle..
203 = 4*50 +3 (since 4 is power cycle freq of 2).
So rem(2^203)/5 changes to rem2^3/5 ie rem8/5 =3.

Rem(3^506)/5 can also be found out by using power cycle.
506=4*126+2 (since 4 is power cycle freq of 3).
So rem(3^506)/5 changes to rem3^2/5 ie rem9/5 =4

Now indiv. rem. are 3 and 4.
Whole remainder is rem(3*4/5) { since the product is greater than devisor,we need to take rem.of the product again.
So answer is rem(12/5) ie 2.


Now rem. of 2^102/3 is equivalent to (rem. 2/3) ^102 that is (-1)^102 =1.

****For finding out the unit place of a very large power of a no. we can use power cycle****



Points to be remembered.
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The frequency of the power cycles of various digits is very important.
This helps you to solve remainder questions involving large powers.
The rem. of a product is the product of the individual rem.
For finding out the unit place of a very large power of a no. we can use power cycle.

Lessons for scoring high marks in Quantitative section in CAT,GMAT...
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Quantitative section in CAT is one of the most dreaded sections by a majority of aspirants. But if we analyze the CAT papers of previous years, Quantitative is one section which can easily increase your overall score and percentile.
In order to increase the score in Quantitative section we need to study Quantitative in a more systematic way. First we need to categorize whole Quantitative portion in to different sections.

Main Categories of Quantitative section
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Numbers
Geometry& Trigonometry
Permutations & Combinations
Progressions
Equations
Functions & Graphs
Algebra
Time & Work
Time & Distance

In a CAT scenario, the sections like Numbers, Geometry , Progressions, P&C are having higher weightage compared to topics like Time & Work, Time & Distance etc. The most important being Numbers. The topic numbers are extensively used for giving questions in both Quantitative and Data Interpretation sections.

Different Approaches to solve Quantitative Questions:-

a.Direct or convectional method of going from the question to the answer.But this method is least preferable in competitive exams like CAT.

b.Elimination of answer options or going with the options.This is the best and quickest way to reach the required answer.

c.Substitution of some values in the question.This method is very much useful for algebra problems

The detailed analysis of each sections and tips for each sections with examples are coming in next posts....