Difference between revisions of "2020 AMC 10A Problems/Problem 22"
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Let <math>a = \left\lfloor \frac{998}n \right\rfloor</math>. Notice that for every integer <math>n \neq 1</math>, if <math>\frac{998}n</math> is an integer, then the three terms in the expression must be <math>(a, a, a)</math>, if <math>\frac{998}n</math> is an integer, then the three terms in the expression must be <math>(a, a + 1, a + 1)</math>, and if <math>\frac{1000}n</math> is an integer, then the three terms in the expression must be <math>(a, a, a + 1)</math>. This is due to the fact that 998, 999, and 1000 share no factors other than 1. | Let <math>a = \left\lfloor \frac{998}n \right\rfloor</math>. Notice that for every integer <math>n \neq 1</math>, if <math>\frac{998}n</math> is an integer, then the three terms in the expression must be <math>(a, a, a)</math>, if <math>\frac{998}n</math> is an integer, then the three terms in the expression must be <math>(a, a + 1, a + 1)</math>, and if <math>\frac{1000}n</math> is an integer, then the three terms in the expression must be <math>(a, a, a + 1)</math>. This is due to the fact that 998, 999, and 1000 share no factors other than 1. | ||
So, there are two cases: | So, there are two cases: | ||
− | <b>Case 1: <math>n</math> divides <math>999</math>< | + | |
+ | |||
+ | <b>Case 1: <math>n</math> divides <math>999</math></b> | ||
+ | |||
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>. | ||
<math>999</math> = <math>3^3 \cdot 37^1</math> | <math>999</math> = <math>3^3 \cdot 37^1</math> | ||
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 <math>n</math> \neq <math>1</math>. | + | However, we have to subtract <math>1</math>, because we have counted <math>1</math> when <math>n</math> \neq <math>1</math>. |
+ | |||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/Ozp3k2464u4 | https://youtu.be/Ozp3k2464u4 |
Revision as of 19:47, 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 998, 999, and 1000 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
when
\neq
.
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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