2019 AMC 10B Problems/Problem 19
- The following problem is from both the 2019 AMC 10B #19 and 2019 AMC 12B #14, so both problems redirect to this page.
Contents
Problem
Let be the set of all positive integer divisors of
How many numbers are the product of two distinct elements of
Solution 1
To find the number of numbers that are the product of two distinct elements of , we first square
and factor it. Factoring, we find
. Therefore,
has
distinct factors. Each of these can be achieved by multiplying two factors of
. However, the factors must be distinct, so we eliminate
and
, as well as
and
, so the answer is
.
Solution by greersc. (Edited by AZAZ12345 and then by greersc once again)
Solution 2
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
. The last two can not be created because the maximum factor of 100,000 involving only 2s or 5s is only
or
. Since the factors chosen must be distinct, the last two numbers cannot be created because those require
or
. 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 (edited by mshell214)
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 10 Problems and Solutions |
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 |
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