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 the case | + | However, we have to subtract <math>1</math>, because we have counted the case <math>n = 1</math> when we already know that <math>n \neq 1</math>. |
<math>8 - 1 = 7</math> | <math>8 - 1 = 7</math> |
Revision as of 19:02, 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 (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
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 1.
Now that we have counted all of the cases, we add them.
, so the answer is
.
~dragonchomper
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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