2024 AMC 12A Problems/Problem 19

The following problem is from both the 2024 AMC 12A #19 and 2024 AMC 10A #22, so both problems redirect to this page.

Problem

Let $a$ , $b$ , and $c$ be pairwise relatively prime positive integers. Suppose one of these numbers is prime, and the other two are perfect squares. If $abc$ has $15a$ divisors and $a^{2}b^{2}c^{2}$ has $15b$ divisors, what is the least possible value of $a + b + c$ ?

$\textbf{(A)}~18\qquad\textbf{(B)}~44\qquad\textbf{(C)}~108\qquad\textbf{(D)}~141\qquad\textbf{(E)}~636$

Solution

Define $\tau{(n)}$ to be the number of divisors of $n$ and let $\nu_{2}{(n)}$ denote the $2$ -adic valuation of $n$ .

Since $a, b, c$ are pairwise relatively prime, the divisor function is multiplicative and $\tau(abc) = \tau(a)\tau(b)\tau(c)$ . Two of these terms will be odd numbers, and the third one will be equal to $2$ , since two of the numbers are perfect squares and one of them is prime. Therefore, $\nu_{2}{(\tau(abc))} = \nu_{2}{(15a)} = \nu_{2}{(a)} = 1$ . If $a$ is a perfect square then $\nu_{2}{(a)}$ is even, and if $a$ is an odd prime, then $\nu_{2}{(a)} = 0$ , so we must have that $a = 2$ . Therefore $a$ is the prime, and $b$ and $c$ are perfect squares.

Now, $\tau(abc) = \tau(a)\tau(b)\tau(c) = 15a = 30 = 2\tau(b)\tau(c)$ . This means that $\tau(b)\tau(c) = 15$ . Since $b, c > 1$ , (if either of them are one, then the other must be at least $3^{14}$ which is clearly not minimum) the only way this is possible is if $\{\tau(a), \tau(b)\} = \{3, 5\}$ . This means that one of them must be the square of a prime, and the other must be the fourth power of a prime. Thus $\tau(a^{2}b^{2}c^{2}) = 3 \cdot 5 \cdot 9 = 15b$ , and $b = 9$ .

Thus $c$ can be any fourth power of a prime that is relatively prime to both $2$ and $9$ , the least of which is $5^4 = 625$ . The least possible value of $a + b + c$ is $2 + 9 + 625 = \boxed{\textbf{(E)}~636}$ .

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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All AMC 12 Problems and Solutions

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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All AMC 10 Problems and Solutions

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2024 AMC 12A Problems/Problem 19

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