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 $n \equiv 1 \pmod{3}$ and $n \equiv 4 \pmod{6}$ , and therefore $n \equiv 3 \pmod{5}$ . Using CRT, 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 $n \equiv 0 \pmod{3}$ , $n \equiv 3 \pmod{6}$ , a contradiction. Thus, $n \equiv 2 \pmod{3}$ , which implies $n \equiv 5 \pmod{6}$ . $n$ can either be $0 \pmod{5}$ , which implies that $n \equiv 35 \pmod{60}$ by CRT, giving $17$ cases; or $4 \pmod{5}$ , which implies that $n \equiv 59 \pmod{60}$ by CRT, giving $16$ cases.

The total number of extra-distinct numbers is thus $16 + 16 + 17 = \boxed{049}$ .

~mathboy100

Solution 2 (Simpler)

Because the LCM of all of the numbers we are dividing by is $60$ , we know that all of the remainders are $0$ again at $60$ , meaning that we have a cycle that repeats itself every $60$ numbers.

After listing all of the remainders up to $60$ , we find that $35$ , $58$ , and $59$ are extra-distinct. So, we have $3$ numbers every $60$ which are extra-distinct. $60\cdot16 = 960$ and $3\cdot16 = 48$ , so we have $48$ extra-distinct numbers in the first $960$ numbers. Because of our pattern, we know that the numbers from $961$ thru $1000$ will have the same remainders as $1$ thru $40$ , so we have $1$ other extra-distinct number ( $35$ ).

$48 + 1 = \boxed{049}$ .

~Algebraik

Solution 3

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

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 $16 + 17 + 16 = \boxed{049}$ .

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

Solution 4 (Small addition to solution 2)

We need to find that $35$ , $58$ , and $59$ are all extra-distinct numbers smaller then $61.$

Let $k \in \left\{2, 3, 4, 5, 6 \right\}.$ Denote the remainder in the division of $a$ by $b$ as ${\rm Rem} \ \left( a, b \right).$

${\rm Rem} \ \left( -1, k \right) = k - 1 \implies {\rm Rem} \ \left( 59, k \right) = k - 1 = \left\{1, 2, 3, 4, 5 \right\}\implies 59$ is extra-distinct. ${\rm Rem} \ \left( -2, k \right) = k - 2 \implies {\rm Rem} \ \left( 58, k \right) = k - 2 = \left\{0, 1, 2, 3, 4 \right\} \implies 58$ is extra-distinct. ${\rm Rem} \ \left( x + 12y, k \right) = {\rm Rem} \ \left(x, k \right) + \left\{0, 0, 0, {\rm Rem} \ \left(12y, k \right), 0 \right\}.$ We need to check all of the remainders up to $12 - 3 = 9$ and remainders ${\rm Rem} \ \left( 59 - 12, k \right) = {\rm Rem} \ \left( 59 - 36, k \right) = \left\{1, 2, 3, 3, 5 \right\},$ ${\rm Rem} \ \left( 59 - 48, k \right) = \left\{1, 2, 3, 1, 5 \right\},$ ${\rm Rem} \ \left( 59 - 24, k \right) ={\rm Rem} \ \left(35, k \right) = \left\{1, 2, 3, 0, 5 \right\} \implies 35$ is extra-distinct.

vladimir.shelomovskii@gmail.com, vvsss

Video Solution

https://youtu.be/8oOik9d1fWM

~MathProblemSolvingSkills.com

2023 AIME I (Problems • Answer Key • Resources)
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