Difference between revisions of "2019 AMC 10B Problems/Problem 19"
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| − | To find the number of numbers that are the product of two distinct elements of <math>S</math>, we first square <math>S</math> and factor it. Factoring, we find <math>S^2 = 2^{10} \cdot 5^{10}</math>. Therefore, <math>S^2</math> has <math>(10 + 1)(10 + 1) = 121</math> distinct factors. Each of these can be achieved by multiplying two factors of <math>S</math>. However, the factors must be distinct, so we eliminate <math>1</math> and <math>S^2</math>, so the answer is <math>121 - | + | To find the number of numbers that are the product of two distinct elements of <math>S</math>, we first square <math>S</math> and factor it. Factoring, we find <math>S^2 = 2^{10} \cdot 5^{10}</math>. Therefore, <math>S^2</math> has <math>(10 + 1)(10 + 1) = 121</math> distinct factors. Each of these can be achieved by multiplying two factors of <math>S</math>. However, the factors must be distinct, so we eliminate <math>1</math> and <math>S^2</math>, 2^10 and 5^10 so the answer is <math>121 - 4 = 117</math>. |
| − | Solution by greersc. | + | Solution by greersc. (Edited by AZAZ12345) |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2019|ab=B|num-b=18|num-a=20}} | {{AMC10 box|year=2019|ab=B|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:41, 14 February 2019
Problem
Let 
be the set of all positive integer divisors of 
How many numbers are the product of two distinct elements of 
Solution
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 , 2^10 and 5^10 so the answer is 
.
Solution by greersc. (Edited by AZAZ12345)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
