Concepts of Co-primes for CAT Exam with Solved Examples and Explanations

Saturday, July 13th, 2019


Concepts-of-Co-primes-for-CAT-Exam-with-Solved-Examples-and-Explanations

Concepts of Co-primes:

Co-primes are those numbers which do not have any common factors between them. For e.g. 4,9,77, are co-primes as they don’t have any common factors.

Note: Co-primes will not always be prime nos. 64 and 27 are co-primes but not prime numbers.

There are three types of questions based on co-primes:

1. Number of ways of writing a number as a product of two co-primes

If N=ap .bq .cr …….., then, the number of ways of writing N as a product of 2 co-primes is 2n-1, where ‘n’ is the number of distinct prime factors of the given number N.
Let’s understood it by a simple question:

Ques: Find the number of ways of writing 48 as a product of two co-primes?

Sol: 48, which is 24 x 31 , the value of ‘n’ is 2 because only 2 distinct prime factors (2&3) are involved.
Hence, the number of ways = 22-1 =2 i.e. 48 can be written as a product of 2 co-primes, in two different ways. They are (1, 48) and (3, 16).

2. Number of co-primes to N, that is less than N

If N is a number that can be written as ap .bq .cr …….., then, the number of co-primes of N, which are less than N, represented by ϕ (N) is,
N(1- 1/a ) (1- 1/b ) (1- 1/c )…………………

For e.g., if 48, is considered,
N= N=ap .bq .cr ……. i.e. 48=24. 31.

Hence, a=2, b=3, p=4, q=1.

ϕ (48) = 48(1- 1/2 ) (1- 1/3 )
= 48 x  1/2  x  1/3
=16

Note: If numbers less than 48 are listed, and co-primes to 48 are spotted, the count of co-primes will be 16.

3. Sum of co-primes to N that less than N

The sum of the co-primes of N, that less than N is N/2. ϕ (N). If we consider the above example, already we have ϕ (48) = 16.

Hence, sum of co-primes of 48 that are less than 48 = N/2. ϕ (N) = 48/2 x 16 = 384.

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Let’s solve some advanced questions on Co-primes:

1. How many integers from 101 to 300 are co-prime to 20?

Sol: 20= 22 x 5

So we need to find the numbers which do not have 2 or 5 as factors.
Numbers between 101 to 300 which are multiple of 2 = 200/2 = 100
Numbers between 101 to 300 which are multiple of 5 = 200/5 = 40

Now we need to find numbers which are multiples of both 2 and 5, as these numbers would be counted twice while counting the numbers divisible by 2 and 5
Numbers between 101 and 300 which are multiple of 10 (both 2 and 5) = 200/10 = 20

So total = 200 – (100 + 40 – 20) = 200 – 120 = 80
Hence, 80 is the answer.

2. How many positive integers between 100 and 200 are co-prime to 630?

Sol: 630 = 2 * 32 * 5 * 7

Numbers between 100 and 200 which are divisible by 2 = 51
Numbers between 100 and 200 which are divisible by 3 = 33
Numbers between 100 and 200 which are divisible by 5 = 21
Numbers between 100 and 200 which are divisible by 7 = 14
Total = 119
Now we need to find out the double counting cases and remove them.

Numbers between 100 and 200 which are divisible
by 2 * 3 = 17
by 2 * 5 = 11
by 2 * 7 = 7
by 3 * 5 = 7
by 3 * 7 = 5
by 5 * 7 = 3
by 2 * 3 * 5 = 3
by 3 * 5 * 7 = 1
by 2 * 5 * 7 = 1
by 2 * 3 * 7 = 2
by 2 * 3 * 5 * 7 = 0
Total (to be removed) = 50 – 7 = 43

So positive integers between 100 and 200 which are co-prime to 630 = 101 – (119 – 43) = 25
Hence, 25 is the answer.

3. In how many ways in which 480 can be written as the product of two numbers which are coprime to each other?

Sol: Formula is 2(n – 1) where n is the distinct primes in the factorized form of the number
480 = 25 * 3 * 5

3 distinct prime factors (2, 3 and 5)
So 480 can be written as the product of 2 co-prime numbers in 2(3 – 1) = 4 ways.

4. How many factor pairs of N = 360 will be co-prime to each other?

Sol: On factorizing 360 we get the following factors:
360 = (23) * (32) * (5)

First, suppose that 5 divides neither co-prime factor. Then we have two cases:

Case 1: One factor must be 2 to some positive power while the other factor must be 3 to some positive power. Hence there are 3 * 2 = 6 of these (as there are 3 possible positive powers of 2 and there are 2 possible positive powers of 3).

Case 2: One factor is 1 while the other factor is a divisor of (2^3) * (3^2).
Hence there are (3+1)(2+1)=12 of these.

Now, note that all of these pairs consist of different numbers except the pair {1,1}. For each of the pairs with two distinct elements, we can multiply either element by 5 to get two new co-prime pairs of factors (that is the solution {a,b} gives rise to the solutions {5a,b}, and {a,5b}.
Finally, the pair {1,1} under this process only gives us one new pair: {1,5}.
In conclusion, we have 3(6 + 12 − 1) + 2 = 53 co-prime pairs.

5. How many three digit numbers less than 200 are co-prime to 638?

Sol: 638 =2 * 11 * 29

3 digit numbers less than 200 divisible by 2 = 50
3 digit numbers less than 200 divisible by 11= 9
3 digit numbers less than 200 divisible by 29=3

Total= 62

Now on taking pairs:

3 digit numbers less than 200 divisible by 2*11=5
3 digit numbers less than 200 divisible by 2*29=2
3 digit numbers less than 200 divisible by 11*29=0

Hence answer is 100- (62-7) = 45

6. If N is a natural number < 100, then for how many values of N are the numbers 6N + 1 and 15N + 2 are relatively prime?

Sol: Suppose 6N + 1 and 15N + 2 are not relatively prime.

Let P is a prime dividing both of them.
Then P also divides 5(6N + 1) and 2(15N + 2) so P divides 30N + 5 and 30N + 4.
Hence P divides the difference (30N + 5) – (30N + 4) = 1.

Now 1 is not divisible by any other number, i.e. it contradicts our conditions.
So the pair 6N + 1 and 15N + 2 are relatively prime for all values of N.

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b) 2 Live Classes (online) every week for doubt clarification
c) Study Material & PDFs for practice and understanding
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e) Previous Year Questions solved on video


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