Difference between revisions of "2020 AMC 10A Problems/Problem 22"
(→Solution 1 (Casework)) |
(→Solution 1 (Casework)) |
||
Line 65: | Line 65: | ||
<math>16 - 1 = 15</math> | <math>16 - 1 = 15</math> | ||
+ | |||
Revision as of 21:30, 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)
Expression:
Solution:
Let .
Notice that for every integer ,
if
is an integer, then the three terms in the expression above must be
,
if
is an integer, then the three terms in the expression above must be
, and
if
is an integer, then the three terms in the expression above must be
.
This is due to the fact that ,
, and
share no common factors other than 1.
Note that doesn't work; to prove this, we just have to substitute
for
in the expression, to get
This gives us which is divisible by 3.
Now, we test the three cases mentioned above.
Case 1: divides
The first case does not work, as the three terms in the expression must be , as mentioned above, so the sum becomes
, which is divisible by
.
Case 2: 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 the case
doesn't work, as mentioned previously.
We now do the same for the third case.
Case 3: divides
=
So, the total number of factors of is
.
Again, we have to subtract , for the reason stated in Case 2.
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.