Difference between revisions of "2023 AIME I Problems/Problem 7"
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Revision as of 16:16, 8 February 2023
Contents
[hide]Problem
Call a positive integer extra-distinct if the remainders when
is divided by
and
are distinct. Find the number of extra-distinct positive integers less than
.
Solution
.
We have . This violates the condition that
is extra-distinct.
Therefore, this case has no solution.
.
We have . This violates the condition that
is extra-distinct.
Therefore, this case has no solution.
.
We have . This violates the condition that
is extra-distinct.
Therefore, this case has no solution.
.
The condition implies
,
.
Because is extra-distinct,
for
is a permutation of
.
Thus,
.
However, conflicts
.
Therefore, this case has no solution.
.
The condition implies
and
.
Because is extra-distinct,
for
is a permutation of
.
Because , we must have
. Hence,
.
Hence, .
Hence,
.
We have .
Therefore, the number extra-distinct
in this case is 16.
.
The condition implies
and
.
Because is extra-distinct,
and
are two distinct numbers in
.
Because
and
is odd, we have
.
Hence,
or 4.
,
,
.
We have .
We have .
Therefore, the number extra-distinct
in this subcase is 17.
,
,
.
.
We have .
Therefore, the number extra-distinct
in this subcase is 16.
Putting all cases together, the total number of extra-distinct is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2
can either be
or
mod
.
Case 1:
Then, , which implies
. By CRT,
, and therefore
. Using CRT again, we obtain
, which gives
values for
.
Case 2:
is then
. If
, then by CRT,
, a contradiction. Thus,
, which by CRT implies
.
can either be
, which implies that
,
cases; or
, which implies that
,
cases.
.
~mathboy100
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.