Question

对 $t = 1, 2, 3, 4$ ，定义 $S_t = \sum_{i=1}^{350} a_i^t$ ，其中 $a_i \in \{1, 2, 3, 4\}$ 。若 $S_1 = 513$ 且 $S_4 = 4745$ ，求 $S_2$ 的最小可能值。

For $t = 1, 2, 3, 4,$ define $S_t = \sum_{i=1}^{350} a_i^t,$ where $a_i \in \{1, 2, 3, 4\}.$ If $S_1 = 513$ and $S_4 = 4745,$ find the minimum possible value for $S_2.$

答案

答案：905

知识点：方程组丢番图方程最优化

难度评级：3060

解答：

对 $j = 1, 2, 3, 4$ ，令 $m_j$ 为等于 $j$ 的 $a_i$ 的个数。则 $\begin{aligned} &m_1 + m_2 + m_3 + m_4 = 350, \\ &m_1 + 2m_2 + 3m_3 + 4m_4 = 513, \\ &m_1 + 16m_2 + 81m_3 \\ &\quad {}+ 256m_4 = 4745. \end{aligned}$

用后两个方程分别减去第一个方程，得到 $m_2 + 2m_3 + 3m_4 = 163$ 和 $15m_2 + 80m_3 + 255m_4 = 4395$ ；后者再减去前者的 $15$ 倍，得到 $50m_3 + 210m_4 = 1950$ ，即 $5m_3 + 21m_4 = 195$ 。因此 $m_4$ 是非负的 $5$ 的倍数，且只有 $m_4 = 0$ 和 $m_4 = 5$ 能使所有变量非负，分别给出 $(m_1, m_2, m_3, m_4)$ $= (226, 85, 39, 0)$ 或 $(215, 112, 18, 5)$ 。

它们分别给出 $S_2 = m_1 + 4m_2$ $+ 9m_3 + 16m_4$ $= 917$ 和 $905$ ，所以最小值为 $905$ 。

For $j = 1, 2, 3, 4,$ let $m_j$ be the number of $a_i$ equal to $j.$ Then $\begin{aligned} &m_1 + m_2 + m_3 + m_4 = 350, \\ &m_1 + 2m_2 + 3m_3 + 4m_4 = 513, \\ &m_1 + 16m_2 + 81m_3 \\ &\quad {}+ 256m_4 = 4745. \end{aligned}$

Subtracting the first equation from the other two gives $m_2 + 2m_3 + 3m_4 = 163$ and $15m_2 + 80m_3 + 255m_4 = 4395;$ subtracting $15$ times the former from the latter leaves $50m_3 + 210m_4 = 1950,$ that is, $5m_3 + 21m_4 = 195.$ Hence $m_4$ is a nonnegative multiple of $5,$ and only $m_4 = 0$ and $m_4 = 5$ keep everything nonnegative, giving $(m_1, m_2, m_3, m_4)$ $= (226, 85, 39, 0)$ or $(215, 112, 18, 5).$

These yield $S_2 = m_1 + 4m_2$ $+ 9m_3 + 16m_4$ $= 917$ and $905$ respectively, so the minimum is $905.$

2009 AIME I 第 14 题

答案：905

难度评级：3060

解答：

其他年份的第 14 题