# CAT Question Bank - Co Primes

• Q20) How many integer pairs (a, b) are possible if LCM (a, b) = 540

• Q21) Let S = { 1, 2, 3, 4, ... 1000}. Let A be subset of S such that any two elements in A are coprime to each other and no element of A is a prime. The maximum possible number of elements in A is
a. 31
b. 30
c. 12
d. 11
e. None of These

• Q22) a, b and c are natural numbers such that a + b + c = 219, what is the maximum possible difference between LCM of any two possible triplets (a, b, c) ?

• Q23) Sum of two numbers is 85 and their LCM is 102. Find the numbers

• Q24) Atleast how many numbers should be taken from the set {1, 2, 3, ..., 50} to ensure that two of the selected numbers are coprime

• Q25) The sum of 6 natural numbers (not necessarily distinct) is 23. Let M denote the positive difference of the maximum and minimum of the LCM of these 6 numbers. Then total number of divisors that divide M^2 but doesn't divide M are
(a) 22
(b) 33
(c) 18
(d) 26

• Q26) How many pairs (x, y) of positive integers with x ≤ y satisfy HCF (x, y) = 5! and LCM (x, y) = 50!

• Q27) How many natural numbers less than 4444 are not - coprime to 247?
a) 585
b) 341
c) 244
d) 557

• Q28) 0.12323232323... can be written in the form of a/b where a and b are integers and a and b are coprime. What is the value of a + b?

• Q29) The product of 3 natural numbers (N1, N2, N3) is 12 times their HCF. Find the number of ordered triplets (N1, N2, N3).

• Q30) 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
a) 16
b) 10
c) 33
d) None of these

• @rowdy-rathore 4
2^(n-1) when n=3

• 20 = 2^2 * 5
So we need to find the numbers which does 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 multiple 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

• 630 = 2 * 3^2 * 5 * 7

Numbers between 100 and 200 which are divisible
by 2 = 51
by 3 = 33
by 5 = 21
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 coprime to 630 = 101 - (119 - 43) = 25

• 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.

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