Let S={1,2,3, …, 2022}. Then the probability, that a randomly chosen number n from the set S such that HCF (n, 2022)=1, is :

Solution

Maths sir · 4 steps

Break 2022 into primes first; then count the numbers that dodge every one of those primes.

1. Factorise 2022. 2022 = 2 × 3 × 337, and 337 is prime because no prime up to 17 divides it and 19² = 361 > 337. So HCF(n, 2022) = 1 means n is divisible by none of 2, 3 and 337.
2. Count with Euler. The number of such n from 1 to 2022 is Euler's totient φ(2022). You remove the fraction divisible by each prime, one prime at a time.

   φ(2022) = 2022(1-1/2)(1-1/3)(1-1/337) = 1 × 2 × 336 = 672

3. Check by inclusion-exclusion. There are 1011 multiples of 2, 674 of 3 and 6 of 337. There are 337 multiples of 6, 3 of 674, 2 of 1011 and 1 of 2022. So 1011 + 674 + 6 - 337 - 3 - 2 + 1 = 1350 numbers share a factor with 2022, and 2022 - 1350 = 672 do not.
4. So. The probability is 672/2022 = 336/1011 = 112/337, which is option (d).

The idea to keep. To count numbers up to N that are coprime to N, factorise N into primes and use φ(N) = N Π (1 - 1/p). Inclusion-exclusion gives the same count and is a good cross-check.

Answer from the NTA final answer key. Solution by PrepWiser.