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>, which is <math>2^10 | + | 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>, which is <math>2^{10}\cdot 5^{10}</math>, so the answer is <math>121 - 4 = 117</math>. |
Solution by greersc. (Edited by AZAZ12345 and then by greersc once again) | Solution by greersc. (Edited by AZAZ12345 and then by greersc once again) |
Revision as of 17:49, 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
, which is
, so the answer is
.
Solution by greersc. (Edited by AZAZ12345 and then by greersc once again)
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.