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 . and do not work since the factors chosen must be distinct, and those require or .
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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