AIME 2023 I · 第 7 题

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 solutions.

$\textbf{Case 4: } {\rm Rem} \ \left( n, 6 \right) = 4$.

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 of 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 of 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 of extra-distinct $n$ in this subcase is 16.

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

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) ~minor formatting changes by PojoDotCom

Solution 4 (Small addition to solution 2)

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

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. $58 - 12 = 46 \implies {\rm Rem} \ \left( 46, 5 \right) = 1 = {\rm Rem} \ \left( 46, 3 \right),$ $58 - 24 = 34 \implies {\rm Rem} \ \left( 34, 5 \right) = 4 = {\rm Rem} \ \left( 34, 6 \right),$ $58 - 36 = 22 \implies {\rm Rem} \ \left( 22, 5 \right) = 2 = {\rm Rem} \ \left( 22, 4 \right),$ $58 - 48 = 10 \implies {\rm Rem} \ \left( 10, 5 \right) = 0 = {\rm Rem} \ \left( 10, 2 \right).$

vvsss

Video Solution

https://youtu.be/8oOik9d1fWM

~MathProblemSolvingSkills.com