*Friday, July 31st, 2020*

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 donâ€™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:

If N=ap .bq .cr â€¦â€¦.., then, the number of ways of writing N as a product of 2 co-primes is 2n-1, where â€˜nâ€™ is the number of distinct prime factors of the given number N.

Letâ€™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 â€˜nâ€™ 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).

If N is a number that can be written as ap .bq .cr â€¦â€¦.., then, the number of co-primes of N, which are less than N, represented by Ï• (N) is,

N(1- 1/a ) (1- 1/b ) (1- 1/c )â€¦â€¦â€¦â€¦â€¦â€¦â€¦

For e.g., if 48, is considered,

N= N=ap .bq .cr â€¦â€¦. i.e. 48=24. 31.

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

Ï• (48) = 48(1- 1/2 ) (1- 1/3 )

= 48 xÂ 1/2Â xÂ 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.

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

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

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Letâ€™s solve some advanced questions on Co-primes:

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

**Sol:** 20= 2^{2 }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 * 3^{2} * 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 = 2^{5} * 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 âˆ’ 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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Quantitative Aptitude – Number Systems – Q1: If the product of three consecutive positive integers is 15600 then the sum of the squares of these integers is

Quantitative Aptitude – Number Systems – Q2: If a, b, c are three positive integers such that a and b are in the ratio 3 : 4 while b and c are in the ratio 2:1, then which one of the following is a possible value of (a + b + c)?

Quantitative Aptitude – Number Systems – Q3: The numbers 1, 2,â€¦,9 are arranged in a 3 X 3 square grid in such a way that each number occurs once and the entries along each column, each row, and each of the two diagonals add up to the same value.

**a)** 900+ Videos covering entire CAT syllabus**b)** 2 Live Classes (online) every week for doubt clarification**c)** Study Material & PDFs for practice and understanding**d)** 10 Mock Tests in the latest pattern**e)** Previous Year Questions solved on video

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