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
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We have . This violates the condition that is extra-distinct. Therefore, this case has no solution.
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We have . This violates the condition that is extra-distinct. Therefore, this case has no solution.
- \textbf{Case 2:} {\rm Rem} \ \left( n, 6 \right) = 2$.
We have$ (Error compiling LaTeX. Unknown error_msg){\rm Rem} \ \left( n, 3 \right) = 2n\textbf{Case 3:} {\rm Rem} \ \left( n, 6 \right) = 3$.
The condition$ (Error compiling LaTeX. Unknown error_msg){\rm Rem} \ \left( n, 6 \right) = 3{\rm Rem} \ \left( n, 2 \right) = 1{\rm Rem} \ \left( n, 3 \right) = 0$.
Because$ (Error compiling LaTeX. Unknown error_msg)n{\rm Rem} \ \left( n, l \right)l \in \left\{ 2, 3, 4 \right\}\left\{ 0, 1 ,2 \right\}{\rm Rem} \ \left( n, 4 \right) = 2$.
However,$ (Error compiling LaTeX. Unknown error_msg){\rm Rem} \ \left( n, 4 \right) = 2{\rm Rem} \ \left( n, 2 \right) = 1\textbf{Case 4:} {\rm Rem} \ \left( n, 6 \right) = 4$.
The condition$ (Error compiling LaTeX. Unknown error_msg){\rm Rem} \ \left( n, 6 \right) = 4{\rm Rem} \ \left( n, 2 \right) = 0{\rm Rem} \ \left( n, 3 \right) = 1$.
Because$ (Error compiling LaTeX. Unknown error_msg)n{\rm Rem} \ \left( n, l \right)l \in \left\{ 2, 3, 4 , 5 \right\}\left\{ 0, 1 ,2 , 3 \right\}$.
Because$ (Error compiling LaTeX. Unknown error_msg){\rm Rem} \ \left( n, 2 \right) = 0{\rm Rem} \ \left( n, 4 \right) = 2{\rm Rem} \ \left( n, 5 \right) = 3$.
Hence,$ (Error compiling LaTeX. Unknown error_msg)n \equiv -2 \pmod{{\rm lcm} \left( 4, 5 , 6 \right)}n \equiv - 2 \pmod{60}$.
We have$ (Error compiling LaTeX. Unknown error_msg)1000 = 60 \cdot 16 + 40n\textbf{Case 5:} {\rm Rem} \ \left( n, 6 \right) = 5$.
The condition$ (Error compiling LaTeX. Unknown error_msg){\rm Rem} \ \left( n, 6 \right) = 5{\rm Rem} \ \left( n, 2 \right) = 1{\rm Rem} \ \left( n, 3 \right) = 2$.
Because$ (Error compiling LaTeX. Unknown error_msg)n{\rm Rem} \ \left( n, 4 \right){\rm Rem} \ \left( n, 5 \right)\left\{ 0, 3, 4 \right\}{\rm Rem} \ \left( n, 4 \right) \leq 3n{\rm Rem} \ \left( n, 4 \right) = 3{\rm Rem} \ \left( n, 5 \right) = 0\textbf{Case 5.1:} {\rm Rem} \ \left( n, 6 \right) = 5{\rm Rem} \ \left( n, 4 \right) = 3{\rm Rem} \ \left( n, 5 \right) = 0$.
We have$ (Error compiling LaTeX. Unknown error_msg)n \equiv 35 \pmod{60}$.
We have$ (Error compiling LaTeX. Unknown error_msg)1000 = 60 \cdot 16 + 40n\textbf{Case 5.2:} {\rm Rem} \ \left( n, 6 \right) = 5{\rm Rem} \ \left( n, 4 \right) = 3{\rm Rem} \ \left( n, 5 \right) = 4n \equiv - 1 \pmod{60}$.
We have$ (Error compiling LaTeX. Unknown error_msg)1000 = 60 \cdot 16 + 40n$in this subcase is 16.
Putting all cases together, the total number of extra-distinct$ (Error compiling LaTeX. Unknown error_msg)n16 + 17 + 16 = \boxed{\textbf{(049) }}$.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)