2021 Fall AMC 12B Problems/Problem 25
Contents
Problem
For a positive integer, let
be the sum of the remainders when
is divided by
,
,
,
,
,
,
,
, and
. For example,
. How many two-digit positive integers
satisfy
Solution 1
Note that we can add to
to get
, but must subtract
for all
. Hence, we see that there are four ways to do that because
. Note that only
is a plausible option, since
indicates
is divisible by
,
indicates that
is divisible by
,
indicates
is divisible by
, and
itself indicates divisibility by
, too. So,
and
is not divisible by any positive integers from
to
, inclusive, except
and
. We check and get that only
and
give possible solutions so our answer is
.
- kevinmathz
Solution 2
Denote by the remainder of
divided by
.
Define
.
Hence,
Hence, this problem asks us to find all , such that
.
:
.
We have .
Therefore, there is no in this case.
:
and
.
The condition implies
.
This further implies
.
Hence,
.
To get , we have
.
However, we have .
Therefore, there is no in this case.
:
for
and
.
The condition implies
with
.
Hence,
and
.
To get , we have
.
However, we have .
Therefore, there is no in this case.
:
for
and
.
To get , we have
.
Hence, we must have and
for
.
Therefore, .
:
for
and
.
The condition implies
with
.
Hence,
and
.
To get , we have
.
However, we have .
Therefore, there is no in this case.
:
for
and
.
To get , we have
.
This can be achieved if ,
,
.
However, implies
. This implies
. Hence,
.
We get a contradiction.
Therefore, there is no in this case.
:
for
and
.
The condition implies
with
.
Hence,
.
To get , we have
. This implies
.
Because and
, we have
.
Hence,
.
However, in this case, we assume
.
We get a contradiction.
Therefore, there is no in this case.
:
for
and
.
To get , we have
. This is infeasible.
Therefore, there is no in this case.
:
for
.
To get , we have
. This is infeasible.
Therefore, there is no in this case.
Putting all cases together, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 3
To get from to
,
would add by
for each remainder
. However, given that some of these remainders can "round down" to
given the nature of mods, we must calculate the possible values of
such that the remainders in
"rounds down" by a total of
, effectively canceling out the adding by
initially.
To do so, we will analyze the "rounding down" for each of :
: subtract by
: subtract by
: subtract by
, but this also implies mod
, so subtract by
.
: subtract by
: subtract by
, but this also implies mod
and
, so subtract by
: too much
: subtract by
: subtract by
, but this also implies mod
and
, so subtract by
: too much
: subtract by
, but this also implies mod
, so subtract by
: too much
: subtract by
: too much
Notice that . By testing these sums, we can easily show that the only time when the total subtraction is
is when
AND
. By CRT,
:
As in solution 1, then, only and
give possible solutions, so our answer is
.
~xHypotenuse
Video Solution
~MathProblemSolvingSkills.com
Video Solution by Mathematical Dexterity
https://www.youtube.com/watch?v=Fy8wU4VAzkQ
2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last problem |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.