2023 AIME I Problems/Problem 7

Revision as of 17:05, 8 February 2023 by Indigo eagle 108 (talk | contribs) (Solution)

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

$\textbf{Case 0:} {\rm Rem} \ \left( n, 6 \right) = 0$.

We have ${\rm Rem} \ \left( n, 2 \right) = 0$. This violates the condition that $n$ is extra-distinct. Therefore, this case has no solution.

$\textbf{Case 1:} {\rm Rem} \ \left( n, 6 \right) = 1$.

We have ${\rm Rem} \ \left( n, 2 \right) = 1$. This violates the condition that $n$ is extra-distinct. Therefore, this case has no solution.

  1. \textbf{Case 2:} {\rm Rem} \ \left( n, 6 \right) = 2$.

We have$ (Error compiling LaTeX. Unknown error_msg){\rm Rem} \ \left( n, 3 \right) = 2$. This violates the condition that$n$is extra-distinct. Therefore, this case has no solution.$\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$implies${\rm Rem} \ \left( n, 2 \right) = 1$,${\rm Rem} \ \left( n, 3 \right) = 0$.

Because$ (Error compiling LaTeX. Unknown error_msg)n$is extra-distinct,${\rm Rem} \ \left( n, l \right)$for$l \in \left\{ 2, 3, 4 \right\}$is a permutation of$\left\{ 0, 1 ,2 \right\}$. Thus,${\rm Rem} \ \left( n, 4 \right) = 2$.

However,$ (Error compiling LaTeX. Unknown error_msg){\rm Rem} \ \left( n, 4 \right) = 2$conflicts${\rm Rem} \ \left( n, 2 \right) = 1$. Therefore, this case has no solution.$\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$implies${\rm Rem} \ \left( n, 2 \right) = 0$and${\rm Rem} \ \left( n, 3 \right) = 1$.

Because$ (Error compiling LaTeX. Unknown error_msg)n$is extra-distinct,${\rm Rem} \ \left( n, l \right)$for$l \in \left\{ 2, 3, 4 , 5 \right\}$is a permutation of$\left\{ 0, 1 ,2 , 3 \right\}$.

Because$ (Error compiling LaTeX. Unknown error_msg){\rm Rem} \ \left( n, 2 \right) = 0$, we must have${\rm Rem} \ \left( n, 4 \right) = 2$. Hence,${\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)}$. Hence,$n \equiv - 2 \pmod{60}$.

We have$ (Error compiling LaTeX. Unknown error_msg)1000 = 60 \cdot 16 + 40$. Therefore, the number extra-distinct$n$in this case is 16.$\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$implies${\rm Rem} \ \left( n, 2 \right) = 1$and${\rm Rem} \ \left( n, 3 \right) = 2$.

Because$ (Error compiling LaTeX. Unknown error_msg)n$is extra-distinct,${\rm Rem} \ \left( n, 4 \right)$and${\rm Rem} \ \left( n, 5 \right)$are two distinct numbers in$\left\{ 0, 3, 4 \right\}$. Because${\rm Rem} \ \left( n, 4 \right) \leq 3$and$n$is odd, we have${\rm Rem} \ \left( n, 4 \right) = 3$. Hence,${\rm Rem} \ \left( n, 5 \right) = 0$or 4.$\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 + 40$. Therefore, the number extra-distinct$n$in this subcase is 17.$\textbf{Case 5.2:} {\rm Rem} \ \left( n, 6 \right) = 5$,${\rm Rem} \ \left( n, 4 \right) = 3$,${\rm Rem} \ \left( n, 5 \right) = 4$.$n \equiv - 1 \pmod{60}$.

We have$ (Error compiling LaTeX. Unknown error_msg)1000 = 60 \cdot 16 + 40$. Therefore, the number extra-distinct$n$in this subcase is 16.

Putting all cases together, the total number of extra-distinct$ (Error compiling LaTeX. Unknown error_msg)n$is$16 + 17 + 16 = \boxed{\textbf{(049) }}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)