2023 AIME I Problems/Problem 7
Unofficial problem statement: Find the number of positive integers from 1 to 1000 that have different mods in mod 2,3,4,5, and 6.
Unofficial Solution: We realize that any such number (mod 2) and (mod 4) must have the same parity, and its values (mod 3) and (mod 6) must have a absolute value difference of 3. Thus the only possibilities for the sequence of mods are 1,2,3,4,5 1,2,3,0,5 and 0,1,2,3,4. Using CRT and summing we get 049.
Solution
\textbf{Case 0:} .
We have . This violates the condition that
is extra-distinct.
Therefore, this case has no solution.
\textbf{Case 1:} .
We have . This violates the condition that
is extra-distinct.
Therefore, this case has no solution.
\textbf{Case 2:} .
We have . This violates the condition that
is extra-distinct.
Therefore, this case has no solution.
\textbf{Case 3:} .
The condition implies
,
.
Because is extra-distinct,
for
is a permutation of
.
Thus,
.
However, conflicts
.
Therefore, this case has no solution.
\textbf{Case 4:} .
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.
\textbf{Case 5:} .
The condition implies
and
.
Because is extra-distinct,
and
are two distinct numbers in
.
Because
and
is odd, we have
.
Hence,
or 4.
\textbf{Case 5.1:} ,
,
.
We have .
We have .
Therefore, the number extra-distinct
in this subcase is 17.
\textbf{Case 5.2:} ,
,
.
.
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)