Question

萨拉本来要把一个两位数和一个三位数相乘，但她漏写了乘号，只把两位数写在三位数左边，形成一个五位数。这个五位数恰好是她本应得到的乘积的九倍。求这两个数的和。

Sarah intended to multiply a two-digit number and a three-digit number, but she left out the multiplication sign and simply placed the two-digit number to the left of the three-digit number, thereby forming a five-digit number. This number is exactly nine times the product Sarah should have obtained. What is the sum of the two-digit number and the three-digit number?

答案

答案：126

知识点：位值丢番图方程整除性

难度评级：2110

解答：

设 $a$ 为两位数， $b$ 为三位数。条件为 $1000a + b = 9ab$ ，整理得 $b(9a - 1) = 1000a$ 。由于 $\gcd(9a - 1, a) = 1$ ，数 $9a - 1$ 必须整除 $1000$ 。

当 $a$ 是两位数时， $9a - 1$ 从 $89$ 到 $890$ ，且 $9a - 1 \equiv 8 \pmod 9$ 。在这个范围内，与 $8$ 模 $9$ 同余的 $1000$ 的唯一因数是 $125$ ，所以 $a = 14$ ，且 $b = \frac{1000 \cdot 14}{125} = 112$ ，确实是三位数。检验： $14112 = 9 \cdot 14 \cdot 112$ 。

所求的和为 $14 + 112 = 126$ 。

Let $a$ be the two-digit number and $b$ the three-digit number. The condition is $1000a + b = 9ab,$ which rearranges to $b(9a - 1) = 1000a.$ Since $\gcd(9a - 1, a) = 1,$ the number $9a - 1$ must divide $1000.$

For a two-digit $a,$ $9a - 1$ runs from $89$ to $890,$ and $9a - 1 \equiv 8 \pmod 9.$ The only divisor of $1000$ in that range congruent to $8$ modulo $9$ is $125,$ giving $a = 14$ and $b = \frac{1000 \cdot 14}{125} = 112,$ which is indeed a three-digit number. Check: $14112 = 9 \cdot 14 \cdot 112.$

The requested sum is $14 + 112 = 126.$

1997 AIME 第 3 题

答案：126

难度评级：2110

解答：

其他年份的第 3 题