2023 IOQM/Problem 9

Revision as of 09:51, 25 October 2023 by Sansgankrsngupta (talk | contribs) (Solution1(Casework))

Problem

Find the number of triples $(a, b, c)$ of positive integers such that (a) $ab$ is a prime;

(b) $bc$ is a product of two primes;

(c) $abc$ is not divisible by square of any prime and

(d) $abc\leq30$

Solution1(Casework)

Since, $ab$ is a prime, this means that one of $a$ and $b$ is 1 and the other is prime. So, there are 2 cases from here:

Case 1($a=1$)

If $a$ is one and $b$ is a prime, this means that $c$ is also a prime but different from $b$ ( as $bc$ is a product of 2 primes but $abc$ is not divisible by the square of any prime)

Now, $abc\leq30$$bc\leq30$, so all possible pairs of $(a,b,c)$ here are $(1,2,3); (1,2,5); (1,2,7); (1,2,11); (1,2,13); (1,3,2); (1,3,5); (1,3,7); (1,5,2); (1,5,3); (1,7,2); (1,7,3); (1,11,2); (1,13,2)$ Total no. of ordered pairs = 14 here

Case 2($b=1$)

If b is one and $a$ is a prime, this means that c is the product of 2 different primes ( as $bc$ is a product of 2 primes but $abc$ is not divisible by the square of any prime)

Now, $abc\leq30$$ac\leq30$ also, $abc$ is not divisible by the square of any prime so a should not divide $c$ so all possible pairs of $(a,b,c)$ here are $(2,1,15); (3,1,10); (5,1,6)$ Total no. of ordered pairs = 3 here

Hence, total no. of triplets $(a,b,c)$ = 14+3= $\boxed{17}$ ~SANSGANKRSNGUPTA