Question

求在一个 $2 \times 3$ 方格的每个格子中填入一个数字的方法数，使得从左到右读出的两个数之和为 $999$ ，从上到下读出的三个数之和为 $99$ 下图是这样的一个例子，因为 $8 + 991 = 999$ ，且 $9 + 9 + 81 = 99$ 。

Find the number of ways to place a digit in each cell of a $2 \times 3$ grid so that the sum of the two numbers formed by reading left to right is $999,$ and the sum of the three numbers formed by reading top to bottom is $99.$ The grid below is an example of such an arrangement because $8 + 991 = 999$ and $9 + 9 + 81 = 99.$

答案

答案：45

知识点：数字位值隔板法

难度评级：2300

解答：

设顶行数字为 $a, b, c$ ，底行数字为 $d, e, f$ 。两个行数相加时，个位数字满足 $c + f \equiv 9 \pmod{10}$ ，而由于 $c + f \le 18$ ，实际上 $c + f = 9$ ，且没有进位。在十位和百位重复同样的论证，得到 $b + e = 9$ 和 $a + d = 9$ 。

三个列数之和为 $10(a + b + c) + (d + e + f)$ $= 99$ 。令 $S = a + b + c$ ，底行数字之和为 $27 - S$ ，所以 $10S + 27 - S = 99$ ，从而 $S = 8$ 。

反过来，任意满足 $a + b + c = 8$ 的数字都会由 $d = 9 - a$ 、 $e = 9 - b$ 、 $f = 9 - c$ ，确定底行，并且两个条件都成立。非负数字解 $a + b + c = 8$ 的个数为 $\binom{10}{2} = 45$ 。

Let the top row hold digits $a, b, c$ and the bottom row $d, e, f.$ In the sum of the two row numbers, the units digits satisfy $c + f \equiv 9 \pmod{10},$ and since $c + f \le 18$ in fact $c + f = 9$ with no carry. Repeating the argument in the tens and hundreds places gives $b + e = 9$ and $a + d = 9.$

The three column numbers add to $10(a + b + c) + (d + e + f)$ $= 99.$ Writing $S = a + b + c,$ the bottom digits sum to $27 - S,$ so $10S + 27 - S = 99$ and $S = 8.$

Conversely, any digits with $a + b + c = 8$ determine the bottom row by $d = 9 - a,$ $e = 9 - b,$ $f = 9 - c,$ and both conditions hold. The number of solutions of $a + b + c = 8$ in nonnegative digits is $\binom{10}{2} = 45.$

2024 AIME II 第 3 题

答案：45

难度评级：2300

解答：

其他年份的第 3 题