Difference between revisions of "2019 AMC 10B Problems/Problem 19"
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==Solution== | ==Solution== | ||
| + | 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 - 1 - 1 = 119</math>. | ||
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| + | Solution by greersc. | ||
==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 15:50, 14 February 2019
Problem
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 , so the answer is 
.
Solution by greersc.
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. 
