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
[hide]Problem
Let be the set of all positive integer divisors of How many numbers are the product of two distinct elements of
Video Solution
https://youtu.be/ZhAZ1oPe5Ds?t=3975
~ pi_is_3.14
Solution
The prime factorization of is . Thus, we choose two numbers and where and , whose product is , where and .
Notice that this is similar to choosing a divisor of , which has divisors. However, some of the divisors of cannot be written as a product of two distinct divisors of , namely: , , , and . The last two cannot be so written because the maximum factor of containing only s or s (and not both) is only or . Since the factors chosen must be distinct, the last two numbers cannot be so written because they would require or . The first two would require and , respectively. This gives candidate numbers. It is not too hard to show that every number of the form , where , and are not both or , 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 |
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