# IIM CAT Preparation Tips

IIM CAT Preparation Tips: October 2010

## Oct 28, 2010

### Number Theory Questions - Factors

Questions:
1. How many numbers are there less than 100 that cannot be written as a multiple of a perfect square greater than 1?
2. Find the smallest number that has exactly 18 factors?

Qn 1: 61 numbers
Qn 2: 180

Qn1 :How many numbers are there less than 100 that cannot be written as a multiple of a perfect square?

To begin with, all prime numbers will be part of this list. There are 25 primes less than 100. (That is a nugget that can come in handy)

Apart from this, any number that can be written as a product of two or more primes will be there on this list. That is, any number of the form pq, or pqr, or pqrs will be there on this list (where p, q, r, s are primes). A number of the form pnq cannot be a part of this list if n is greater than 1, as then the number will be a multiple of p2.

This is a brute-force question.

First let us think of all multiples of 2 * prime number. This includes 2*3, 2*5, 2*7, 2 * 11 all the way up to 2*47 (14 numbers)

The, we move on to all numbers of the type 3 * prime number 3*5, 3*7 all the way up to 3*31 (9 numbers)

Then, all numbers of the type 5 * prime number – 5*7, 5*11, 5*13, 5*17, 5*19 (5 numbers),

Then, all numbers of the type 7 * prime number  and then 7*11, 7*13 (2 numbers).

There are no numbers of the form 11 * prime number which have not been counted earlier.

Post this, we need to count all numbers of the form p*q*r, where p, q, r are all prime.
In this list, we have 2*3*5, 2*3*7, 2*3*11, 2*3*13 and 2*5*7. Adding 1 to this list, we get totally 36 different composite numbers.

Along with the 25 prime numbers, we get 61 numbers that cannot be written as a product of a perfect square greater than 1

Alternative Method

There is another method of solving this question.

We can list all multiples of perfect squares (without repeating any number) and subtract this from 99

4 - there are 24 multiples of 4 { 4, 8, 12, …96}
9 - There are 11 multiples, 2 are common with 4 (36 and 72), so let us add 9 new numbers to the list{ 9, 18, 27, ….99}
16 - 0 new multiples
25 - 3 new multiples { 25, 50, 75
36 – 0 new ones
49 – 2 { 49, 98}
64 - 0
81 - 0

So, total multiples of perfect squares are 38. There are 99 numbers totally. So, there are 61 numbers that are not multiples of perfect squares

This is a difficult and time-consuming question. But a question that once solved, helps practice brute-force counting. Another takeaway is the fact that there are 25 primes less than 100. There is a function called pi(x) that gives the number of primes less than or equal to x. pi(10) = 4, pi(100) = 25

4. Find the smallest number that has exactly 18 factors?

Any number of the form paqbrc will have (a+1) (b+1)(c+1) factors, where p, q, r are prime. (This is a very important idea)
Now, the number we are looking for has 18 factors. It can comprise one prime, two primes or three primes.
Now, 18 can be written as 1 * 18 or 3 * 6 or 9 * 2 or 2 * 3 * 3

If we take the underlying prime factorization of N to be paqb, then it can be of the form
p1q8 or p2 q5

If we take the underlying prime factorization of N to be pa, then it can be of the form
p17
If we take the underlying prime factorization of N to be paqbrc, then it can be of the form
p1q1r2
So, N can be of the form p17, p2q5, p1q8 or p1q2r2

Importantly, these are the only possible prime factorizations that can result in a number having 18 factors.

Now, let us think of the smallest possible number in each scenario
p17 - Smallest number = 217
p2q5 – 32 * 25
p1q8 – 31 * 28
p1q1r2 – 51 * 31 * 22
The smallest of these numbers is 51 * 31 * 22  = 180

Labels: ,

## Oct 26, 2010

### Number Theory - Factors Questions

Number Theory is one of the heavily tested topics in CAT and is probably best to practice a lot of questions in this topic.

This is a very interesting topic to prepare for and a fun topic if you go about it the right way. I have given two questions on Number Theory – on the topic of factors below.

Questions
1. A number N^2 has 15 factors. How many factors can N have?
2. If a three digit number ‘abc’ has 2 factors (where a, b, c are digits), how many factors does the 6-digit number ‘abcabc’ have?

Question 1: 6 or 8 factors
Question 2: 16 factors

Qn: A number N^2 has 15 factors. How many factors can N have?

Any number of the form paqbrc will have (a+1) (b+1)(c+1) factors, where p, q, r are prime. (This is a very important idea)
N2 has 15 factors.
Now, 15 can be written as 1 * 15 or 3 * 5.
If we take the underlying prime factorization of N2 to be paqb, then it should have (a + 1) (b+1) factors. So, N can be of the form

p14 or p2q4
p14 will have (14+1) = 15 factors
p2q4 will have (2+1) * (4 +1) = 15 factors.

Importantly, these are the only two possible prime factorizations that can result in a number having 15 factors.

Qn: If a three digit number ‘abc’ has 2 factors (where a, b, c are digits), how many factors does the 6-digit number ‘abcabc’ have?

To start with ‘abcabc’ = ‘abc’ *1001 or abc * 7*11*13 (This is a critical idea to remember)
‘abc’ has only two factors. Or, ‘abc’ has to be prime. Only a prime number can have exactly two factors. (This is in fact the definition of a prime number)

So, ‘abcabc’ is a number like 101101 or 103103.
’abcabc’ can be broken as ‘abc’ * 7 * 11 * 13. Or, a p * 7 * 11 * 13 where p is a prime.

As we have already seen, any number of the form paqbrc will have (a+1) (b+1)(c+1) factors, where p, q, r are prime.

So, p * 7 * 11 * 13 will have = (1+1)*(1+1)*(1+1)*(1+1) = 16 factors

Labels: , , ,

## Oct 25, 2010

### Solutions to final 4 questions from challenge quiz

I have given below the solutions to the final four questions from the challenge quiz. This completes the challenge quiz discussion. Please feel free to send feedback on this.

The solutions can be found here

## Oct 20, 2010

### Discussion on Number Theory questions from Challenge quiz 1

Given below are solutions to three questions that were from Number Theory in our challenge quiz .

The first question is the most interesting as it tests some fundamental properties in Number Theory. The second and third questions can be done by trial and error.

Labels: ,

## Oct 19, 2010

### Challenge quiz - Three questions from Number Theory

Hi all,

I have given below three questions from Number Theory from the quiz that is on our website (We have already discussed three out of these). The questions can be found here - . I will post the solutions for this in a day or two.

Happy cracking.

1. What is the smallest number that has exactly 12 factors?

a. 211
b. 60
c. 120
d. 96
e. 144

2. If f(x) is the number of primes less than or equal to x, find the value of f(90) - f(80) = ?

a. 3
b. 2
c. 1
d. 4
e. 0

3. The difference between the ages of two brothers is a prime number. Sum of their ages is also a prime number. If the elder brother is 28 years old, how many different values can the age of the younger brother take?

a. 2
b. 3
c. 4
d. 1
e. 0

Labels: , ,

## Oct 18, 2010

### CAT Challenge Qn -1. Mensuration Solution

The solution for the question on mensuration is given in the slideshow below. This is a fairly challenging question on mensuration. The question involves breaking down the given region into simpler geometric shapes and then finding the overlapping area.

## Oct 15, 2010

### CAT Challenging Question Mensuration

Here is an interesting question on mensuration

Question

Two horses are tied to opposite vertices of a rectangular field of length and breadth 6 and 6√2 meters. If they are tied with rope of length 6 meters, what is the common area that both horses can graze?
1. 6 π - 9√3
2. 12 π - 18√3
3. π - 2√3
4. 16 π -8√3
5. Cannot be calculated

People think that questions in 3-d (Cubes, Cuboid, cylinder, sphere etc) are more difficult in mensuration. In my view, the best questions are still from the 2-dimensional shapes. In 3-d, once you know the formula, you can pretty much solve everything. In 2-d, questions can include geometry and mensuration. Those are the best ones.

Anyways, happy solving. Will post solutions soon enough.

Cheers,
2iim-CAT classes, correspondence material
Next CAT weekend batch @ Chennai starts Oct 23rd, 2010

## Oct 13, 2010

### CAT - questions and solution from progressions

Have given one interesting question from progressions. This question is from challenge quiz I on our website. This is not that difficult, but as with all good questions, there is a formula-based method of solving this and there is a more straightforward intuitive way of solving them

Question

Sum of first 12 terms of a GP is equal to the sum of the first 14 terms in the same GP. Sum of the first 17 terms is 92, what is the third term in the GP?
1. 92
2. -92
3. 46
4. 231
Correct Answer : 92. Correct Choice : (1)

Sum of first 12 terms is equal to sum of first 14 terms.
Sum of first 14 terms = Sum of first 12 terms + 13th term + 14th term
=> 13th term + 14th term = 0

Let us assume 13th term = k, common ratio = r. 14th term will be kr.
k + kr = 0
k (1 + r) = 0
=> r = -1 as k cannot be zero

Common ratio = -1.
Now, if the first term of this GP is a, second term would be -a, third would be a and so on
The GP would be a, -a, a, -a, a, -a,...

Sum to even number of terms = 0
Sum to odd number of terms = a

Sum to 19 terms is 92 => a = 92
Third term = a = 92

-------------------------------------------------------------------------
2iim - CAT Classes, Correspondence material.
Next CAT weekend batch @ Chennai Starts Oct 23rd, 2010

## Oct 12, 2010

### CAT - questions and solution from progressions

Have given one interesting question from progressions. This is from challenge quiz I on our website. This  is not that difficult, but as with all good questions, there is a formula-based method of solving them and there is a more straightforward intuitive way of solving them. (This question can be solved without putting pen to paper. Try it)

Question

Sum of first 25 terms in AP is 525, sum of the next 25 terms is 725, what is the common difference?
1. 8/25
2. 4/25
3. 6/25
4. 1/5
Correct Answer : 8/25. Correct Choice : (1)

The sum of the first 25 terms, S25 = 525
The sum of the next 25 terms, K25 = sum of next 25 terms = 725
a26 = a1 + 25d
a27 = a2 + 25d

K25 = a26 + a27 + ...... + a50
Or K25 = a1 + 25d + a2 + 25d + ...... + a25+ 25d
Or K25 = a1 + a2 + ...... + a25+ 25(25d)

Or K25 = S25 + 25(25d)
i.e., 725 = 525 + 25 x 25d
200 = 25 x 25d
8 = 25d
d = 8/25

-------------------------------------------------------------------------
2iim - CAT Classes, Correspondence material.
Next CAT weekend batch @ Chennai Starts Oct 23rd, 2010

## Oct 11, 2010

### Solution for CAT Geometry Question 1

The solution to the geometry question on in-radius is given below. The question can be found here . This is a reasonably tough question and quite possibly in the higher end of difficulty level of questions that appear in the CAT.

One of our students has provided another solution for this using trigonometry as well. If you can get that solution, drop a comment. I will be more than happy to add that also on to the blog.

Labels:

## Oct 9, 2010

### CAT Geometry Question

Here is a good geometry question
The unequal side of an isosceles triangle is thrice the inradius of the triangle. What is the ratio of the longest side to the shortest side of this triangle?

(A) 1.5 : 1
(B) 1.3 : 1
(C) 1.2: 1
(D) 1.6 : 1

Solution for this question.

2iim-CAT classes, correspondence material
Next CAT weekend batch @ Chennai starts Oct 23rd, 2010