1990 AIME Problems/Problem 5

Problem

Let $n^{}_{}$ be the smallest positive integer that is a multiple of $75_{}^{}$ and has exactly $75_{}^{}$ positive integral divisors, including $1_{}^{}$ and itself. Find $n/75^{}_{}$ .

Solution

The prime factorization of $75 = 3^15^2$ . Thus, for $n$ to have exactly $75$ integral divisors, we need to have $\displaystyle n = a^{x-1}b^{y-1}\ldots$ such that $x \cdot y \cdot \ldots = 75$ . Since we know that $n$ is divisible by $75$ , two of the prime factors must be $3$ and $5$ . To minimize $n$ , a third factor which is less than $5$ can be used; the only possible prime number is $2$ . Also to minimize $n$ , we want $5$ , the greatest of all the factors, to be raised to the least power. Therefore, $n = 2^43^45^2$ and $\frac{n}{75} = \frac{2^43^45^2}{3 \cdot 5^2} = 16 \cdot 27 = 432$ .

1990 AIME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1990 AIME Problems/Problem 5

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