AIME 1993 · 第 9 题

Solution

The label $1993$ will occur on the $\frac12(1993)(1994) \pmod{2000}$th point around the circle. (Starting from 1) A number $n$ will only occupy the same point on the circle if $\frac12(n)(n + 1)\equiv \frac12(1993)(1994) \pmod{2000}$.

Simplifying this expression, we see that $(1993)(1994) - (n)(n + 1) = (1993 - n)(1994 + n)\equiv 0\pmod{2000}$. Therefore, one of $1993 - n$ or $1994 + n$ is odd, and each of them must be a multiple of $125$ or $16$.

For $1993 - n$ to be a multiple of $125$ and $1994 + n$ to be a multiple of $16$, $n \equiv 118 \pmod {125}$ and $n\equiv 6 \pmod {16}$. The smallest $n$ for this case is $118$.

In order for $1993 - n$ to be a multiple of $16$ and $1994 + n$ to be a multiple of $125$, $n\equiv 9\pmod{16}$ and $n\equiv 6\pmod{125}$. The smallest $n$ for this case is larger than $118$, so $\boxed{118}$ is our answer.

Note: One can just substitute $1993\equiv-7\pmod{2000}$ and $1994\equiv-6\pmod{2000}$ to simplify calculations.

Solution 2

Two labels $a$ and $b$ occur on the same point if $\ a(a+1)/2\equiv \ b(b+1)/2\pmod{2000}$. If we assume the final answer be $n$, then we have $\frac12(n)(n + 1)\equiv \frac12(1993)(1994) \pmod{2000}$.

Multiply $2$ on both side we have $(1993)(1994) - (n)(n + 1) = (1993 - n)(1994 + n)\equiv 0\pmod{4000}$. As they have different parities, the even one must be divisible by $32$. As $(1993 - n)+(1994 + n)\equiv 2\pmod{5}$, one of them is divisible by $5$, which indicates it's divisible by $125$.

Which leads to four different cases: $1993-n\equiv 0\pmod{4000}$ ; $1994+n\equiv 0\pmod{4000}$ ; $1993-n\equiv 0\pmod{32}$ and $1994+n\equiv 0\pmod{125}$ ; $1993-n\equiv 0\pmod{125}$ and $1994+n\equiv 0\pmod{32}$. Which leads to $n\equiv 1993,2006,3881$ and $118\pmod{4000}$ respectively, and only $n=118$ satisfied.Therefore answer is $\boxed{118}$.(by ZJY)