Difference between revisions of "2020 AMC 10A Problems/Problem 22"
(→Solution 2 (Casework)) |
(→Solution 1) |
||
Line 4: | Line 4: | ||
<math>\textbf{(A) } 22 \qquad\textbf{(B) } 23 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 26</math> | <math>\textbf{(A) } 22 \qquad\textbf{(B) } 23 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 26</math> | ||
− | |||
− | |||
− | |||
== Solution 2 (Casework) == | == Solution 2 (Casework) == |
Revision as of 19:01, 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 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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.