Difference between revisions of "2020 AMC 10A Problems/Problem 22"
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So, the total number of factors of <math>999</math> is <math>4 \cdot 2 = 8</math>. | So, the total number of factors of <math>999</math> is <math>4 \cdot 2 = 8</math>. | ||
− | However, we have to subtract <math>1</math>, because we have counted <math>1</math> when we already know that <math>n | + | However, we have to subtract <math>1</math>, because we have counted <math>1</math> when we already know that <math>n \neq 1</math>. |
==Video Solution== | ==Video Solution== |
Revision as of 18:50, 1 February 2020
Problem
For how many positive integers isnot divisible by ? (Recall that is the greatest integer less than or equal to .)
Solution 1
Let . If the expression is not divisible by , then the three terms in the expression must be , which would imply that is a divisor of but not , or , which would imply that is a divisor of but not . has factors, and has factors. However, does not work because a divisor of both and , and since is counted twice, the answer is .
Solution 2 (Casework)
Let . Notice that for every integer , if is an integer, then the three terms in the expression must be , if is an integer, then the three terms in the expression must be , and if is an integer, then the three terms in the expression must be . This is due to the fact that , , and share no factors other than 1. So, there are two cases:
Case 1: n divides 999
Because divides , the number of possibilities for is the same as the number of factors of , excluding .
=
So, the total number of factors of is . However, we have to subtract , because we have counted when we already know that .
Video Solution
~IceMatrix
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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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