Competition problem — Number theory

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Number theory Difficulty 4.5 AIME Find the answer Philippines

Problem:

Let $m$ be the product of all positive integer divisors of $360{,}000$ . Suppose the prime factors of $m$ are $p_{1}, p_{2}, \ldots, p_{k}$ , for some positive integer $k$ , and $m = p_{1}^{e_{1}} p_{2}^{e_{2}} \cdot \ldots \cdot p_{k}^{e_{k}}$ , for some positive integers $e_{1}, e_{2}, \ldots, e_{k}$ . Find $e_{1} + e_{2} + \ldots + e_{k}$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solution:

Let $d(m)$ be the number of positive divisors of $m$ . Since $360{,}000 = 2^{6} \cdot 3^{2} \cdot 5^{4}$ , we have $m = 360{,}000^{\frac{d(m)}{2}}$ . Thus,
$e_{1} + e_{2} + e_{3} = \frac{(6 + 2 + 4)}{2} \cdot [(6 + 1)(2 + 1)(4 + 1)] = 630$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.