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
Let denote the product of the distinct proper divisors of
. A number
is nice in one of two instances:
- It has exactly two distinct prime divisors.
- If we let
, where
are the prime factors, then its proper divisors are
and
, and
.
- If we let
- It is the cube of a prime number.
- If we let
with
prime, then its proper divisors are
and
, and
.
- If we let
We now show that the above are the only two cases. Suppose that another nice number existed that does not fall into one of these two categories. Then we can either express it in the form (with
prime and
) or
(with
).
In the former case, it suffices to note that . In the latter case, then
. For
, we need
(the case
does not work).
Thus, listing out the first ten numbers to fit this form,
.
Summing these yields
.
See also
1987 AIME (Problems • Answer Key • Resources) | ||
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Followed by Problem 4 | |
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