Difference between revisions of "2005 AMC 10A Problems/Problem 21"
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But <math>0</math> isn't a positive integer, so only <math>1</math>, <math>2</math>, <math>3</math>, <math>5</math>, and <math>11</math> are possible values of <math>n</math>. Therefore the number of possible values of <math>n</math> is <math>5\Longrightarrow \boxed{\mathrm{(B)}}</math>. | But <math>0</math> isn't a positive integer, so only <math>1</math>, <math>2</math>, <math>3</math>, <math>5</math>, and <math>11</math> are possible values of <math>n</math>. Therefore the number of possible values of <math>n</math> is <math>5\Longrightarrow \boxed{\mathrm{(B)}}</math>. | ||
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+ | CHECK OUT Video Solution: https://youtu.be/WRv86DHa3zY | ||
==See also== | ==See also== |
Revision as of 20:51, 30 October 2020
Problem
For how many positive integers does evenly divide from ?
Solution
If evenly divides , then is an integer.
Since we may substitute the RHS in the above fraction. So the problem asks us for how many positive integers is an integer, or equivalently when for a positive integer .
is an integer when is a factor of .
The factors of are , , , , , and , so the possible values of are , , , , , and .
But isn't a positive integer, so only , , , , and are possible values of . Therefore the number of possible values of is .
CHECK OUT Video Solution: https://youtu.be/WRv86DHa3zY
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 |
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