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 $n = a^{(5 - 1)}b^{(5 - 1)}c^{(3 - 1)}$ . Since we know that $n$ is divisible by $75$ , two of the factors must be $3$ and $5$ . To minimize $n$ , the last factor must be $2$ ; also to minimize it, we want $5<math> to be raised to the least power. Therefore, <math>n = 2^43^45^2$ and $\frac{n}{75} = \frac{2^43^45^2}{35^2} = 16 \cdot 27 = 432$ .

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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