2023 AIME I Problems/Problem 7

Problem

Call a positive integer $n$ extra-distinct if the remainders when $n$ is divided by $2, 3, 4, 5,$ and $6$ are distinct. Find the number of extra-distinct positive integers less than $1000$ .

Solution 1

$n$ can either be $0$ or $1$ mod $2$ .

Case 1: $n \equiv 0 \pmod{2}$

Then, $n \equiv 2 \pmod{4}$ , which implies $n \equiv 1 \pmod{3}$ . By CRT, $n \equiv 4 \pmod{6}$ , and therefore $n \equiv 3 \pmod{5}$ . Using CRT again, we obtain $n \equiv 58 \pmod{60}$ , which gives $16$ values for $n$ .

Case 2: $n \equiv 1 \pmod{2}$

$n$ is then $3 \pmod{4}$ . If $n \equiv 0 \pmod{3}$ , then by CRT, $n \equiv 3 \pmod{6}$ , a contradiction. Thus, $n \equiv 2 \pmod{3}$ , which by CRT implies $n \equiv 5 \pmod{6}$ . $n$ can either be $0 \pmod{5}$ , which implies that $n \equiv 35 \pmod{60}$ , $17$ cases; or $4 \pmod{5}$ , which implies that $n \equiv 59 \pmod{60}$ , $16$ cases.

$16 + 16 + 17 = \boxed{49}$ .

~mathboy100

Solution 2

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

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.

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

We have ${\rm Rem} \ \left( n, 3 \right) = 2$ . 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 ${\rm Rem} \ \left( n, 6 \right) = 3$ implies ${\rm Rem} \ \left( n, 2 \right) = 1$ , ${\rm Rem} \ \left( n, 3 \right) = 0$ .

Because $n$ 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, ${\rm Rem} \ \left( n, 4 \right) = 2$ 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 ${\rm Rem} \ \left( n, 6 \right) = 4$ implies ${\rm Rem} \ \left( n, 2 \right) = 0$ and ${\rm Rem} \ \left( n, 3 \right) = 1$ .

Because $n$ 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 ${\rm Rem} \ \left( n, 2 \right) = 0$ , we must have ${\rm Rem} \ \left( n, 4 \right) = 2$ . Hence, ${\rm Rem} \ \left( n, 5 \right) = 3$ .

Hence, $n \equiv -2 \pmod{{\rm lcm} \left( 4, 5 , 6 \right)}$ . Hence, $n \equiv - 2 \pmod{60}$ .

We have $1000 = 60 \cdot 16 + 40$ . Therefore, the number extra-distinct $n$ in this case is 16.

$\textbf{Case 5: } {\rm Rem} \ \left( n, 6 \right) = 5$ .

The condition ${\rm Rem} \ \left( n, 6 \right) = 5$ implies ${\rm Rem} \ \left( n, 2 \right) = 1$ and ${\rm Rem} \ \left( n, 3 \right) = 2$ .

Because $n$ 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 $n \equiv 35 \pmod{60}$ .

We have $1000 = 60 \cdot 16 + 40$ . 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 $1000 = 60 \cdot 16 + 40$ . Therefore, the number extra-distinct $n$ in this subcase is 16.

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

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

2023 AIME I (Problems • Answer Key • Resources)
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2023 AIME I Problems/Problem 7

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Problem

Solution 1

Solution 2

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