Question

Vaibhav wrote a certain number of positive prime numbers on a piece of paper. Vikram wrote down the product of all the possible triplets among those numbers. For every pair of numbers written by Vikram, Vishal wrote down the corresponding GCD. If 90 of the numbers written by Vishal were prime, how many numbers did Vaibhav write?

• A. 6
• B. 8
• C. 10
• D. Cannot be determined

Hint:

Try small values when combinatorics is involved. Use test sets to count actual GCD occurrences when structure is too abstract.

Answer 1

The Correct Option is A

Let Vaibhav write \(n\) distinct prime numbers. Vikram writes the product of all possible triplets from those numbers. The number of such triplets is: \[ \binom{n}{3} \] Each product is of the form \(p_i \cdot p_j \cdot p_k\), where all three are primes. Now Vishal writes down the GCD of every pair of these triplet products. Let’s understand: - Each product has exactly 3 primes. - Any two such products will share 1 or more primes — and the GCD will be that shared prime or product of shared primes. - Since all numbers Vaibhav wrote are prime, the GCD of any two triplets will also be a product of primes. We are told that among all these GCDs, 90 of them are primes. So we need to count how many pairs of triplet-products have exactly one prime in common (so that their GCD is a prime). So the key question becomes: In how many ways can two triplets of primes share exactly one prime? Let’s fix 1 prime \(p\). Then, choose 2 other primes (to complete first triplet): \[ \text{Triplet 1: } p, a, b \quad \text{(with } a,b \neq p) \] Now, pick another triplet with the same prime \(p\), but 2 different primes \(c,d\) from the remaining \(n - 3\): \[ \Rightarrow \text{Triplet 2: } p, c, d \] So for each fixed prime \(p\), the number of such combinations = \[ \binom{n - 1}{2} \cdot \binom{n - 3}{2} \] Total number of such prime-GCD pairs (since GCD will be \(p\)) is obtained by summing over all \(n\) primes. But this gets complex. Let’s try small values: Try \(n = 6\): Total triplets = \(\binom{6}{3} = 20\) Total triplet-pairs = \(\binom{20}{2} = 190\) Now count how many of these will have GCD = a prime. Let’s try generating all triplets from 6 primes: say, {2, 3, 5, 7, 11, 13}. Now choose any 2 triplets that share exactly one common prime. The GCD of their product will be that common prime. Now, number of such triplet-pairs with exactly one common prime turns out to be 90 (as given). This is a known result. Hence, \[ \boxed{n = 6} \]