Difference between revisions of "2020 AMC 10A Problems/Problem 22"
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− | <b>Case 1:</b> <math>n</math> divides 999 | + | <b>Case 1:</b> <math>n</math> divides <math>999</math> |
Because <math>n</math> divides <math>999</math>, the number of possibilities for <math>n</math> is the same as the number of factors of <math>999</math>, excluding <math>1</math>. | Because <math>n</math> divides <math>999</math>, the number of possibilities for <math>n</math> is the same as the number of factors of <math>999</math>, excluding <math>1</math>. | ||
Line 20: | Line 20: | ||
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 the case where <math>n = 1</math> when we already know that <math>n \neq 1</math>. | However, we have to subtract <math>1</math>, because we have counted the case where <math>n = 1</math> when we already know that <math>n \neq 1</math>. | ||
+ | |||
+ | <math>8 - 1 = 7</math> | ||
+ | |||
+ | We now do the same for the second case. | ||
+ | |||
+ | |||
+ | |||
+ | <b>Case 2:</b> <math>n</math> divides <math>1000</math> | ||
+ | |||
+ | <math>1000</math> = <math>5^3 \cdot 2^3</math> | ||
+ | |||
+ | So, the total number of factors of <math>1000</math> is <math>4 \cdot 4 = 16</math>. | ||
+ | |||
+ | Again, we have to subtract <math>1</math>, for the reason mentioned above in Case !. | ||
+ | |||
+ | <math>16 - 1 = 15</math> | ||
+ | |||
+ | |||
+ | |||
+ | Now that we have counted all of the cases, we add them. | ||
+ | |||
+ | <math>7 + 15 = 22</math>, so the answer is | ||
==Video Solution== | ==Video Solution== |
Revision as of 18:58, 1 February 2020
Problem
For how many positive integers is
not 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: divides
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 the case where
when we already know that
.
We now do the same for the second case.
Case 2: divides
=
So, the total number of factors of is
.
Again, we have to subtract , for the reason mentioned above in Case !.
Now that we have counted all of the cases, we add them.
, so the answer is
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.