1987 AIME Problems/Problem 3

Problem

By a proper divisor of a natural number we mean a positive integral divisor other than 1 and the number itself. A natural number greater than 1 will be called "nice" if it is equal to the product of its distinct proper divisors. What is the sum of the first ten nice numbers?

Solution

A number is nice in one of two instances: (1) it has exactly two distinct prime divisors (excluding $1$ and itself, note that squares like $3^2 = 9$ don't work, an example would be $3 \cdot 5 = 15$ ), or (2) it the cube of a prime number (an example would be $2^3 = 8$ ). Thus, listing them all out, we get:

$2 \cdot 3 = 6,\ 2^3 = 8,\ 2 \cdot 5 = 10,\ 2 \cdot 7 = 14,\ 3 \cdot 5 = 15,\ 3 \cdot 7 = 21,\ 2 \cdot 11 = 22,\ 2 \cdot 13 = 26,\ 3^3 = 27,\ 3 \cdot 11 = 33$

Directly summing them up yields $2(3 + 4 + 5 + 7 + 11 + 13) + 3(5 + 7 + 9 + 11) = 182$ .

1987 AIME (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1987 AIME Problems/Problem 3

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