Difference between revisions of "2019 AMC 10B Problems/Problem 19"
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Revision as of 03:30, 17 February 2019
- The following problem is from both the 2019 AMC 10B #19 and 2019 AMC 12B #14, so both problems redirect to this page.
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 
. 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 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
