2019 AMC 12B Problems/Problem 14
Revision as of 14:40, 14 February 2019 by Scrabbler94 (talk | contribs) (previous answer was incorrect)
Problem
Let be the set of all positive integer divisors of
How many numbers are the product of two distinct elements of
Solution
The prime factorization of 100,000 is . Thus, we choose two numbers
and
where
and
, whose product is
, where
and
.
Consider . The number of divisors is
. However, some of the divisors of
cannot be written as a product of two distinct divisors of
, namely:
,
,
, and
. This gives
candidate numbers. It is not too hard to show that every number of the form
where
, and
are not both 0 or 10, can be written as a product of two distinct elements in
. Hence the answer is
.
-scrabbler94
See Also
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |