Difference between revisions of "2023 AIME I Problems/Problem 7"
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− | + | ==Problem== | |
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+ | Call a positive integer <math>n</math> extra-distinct if the remainders when <math>n</math> is divided by <math>2, 3, 4, 5,</math> and <math>6</math> are distinct. Find the number of extra-distinct positive integers less than <math>1000</math>. | ||
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==Solution== | ==Solution== | ||
Revision as of 17:11, 8 February 2023
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
is
, then by CRT,
, a contradiction. Thus,
, which by CRT implies
.
can either be
, which implies that
,
cases; or
, which implies that
,
cases.
.
~mathboy100