Question

已知 $\log_{6} a + \log_{6} b + \log_{6} c = 6$ ，其中 $a$ 、 $b$ 、 $c$ 是正整数，它们组成递增等比数列，且 $b - a$ 是某个整数的平方。求 $a + b + c.$

It is given that $\log_{6} a + \log_{6} b + \log_{6} c = 6,$ where $a,$ $b,$ and $c$ are positive integers that form an increasing geometric sequence and $b - a$ is the square of an integer. Find $a + b + c.$

答案

答案：111

知识点：对数等比数列整除性

难度评级：2170

解答：

对数相加得 $\log_6(abc) = 6$ ，所以 $abc = 6^6$ 。等比数列满足 $ac = b^2$ ，于是 $b^3 = 6^6$ ，从而 $b = 36$ ，并且 $ac = 36^2 = 1296$ 。

因为数列递增， $b - a$ 是正平方数，所以 $a = 36 - k^2$ ，其中 $k = 1, \ldots, 5$ 。候选值为 $35, 32, 27, 20, 11$ 。同时 $a$ 必须整除 $1296 = 2^4 \cdot 3^4$ ，候选值中只有 $27$ 满足，此时 $c = 1296/27 = 48$ 。

的确， $27, 36, 48$ 是公比为 $\frac{4}{3}$ 的等比数列，所以 $a + b + c = 27 + 36 + 48 = 111$ 。

Adding the logs gives $\log_6(abc) = 6,$ so $abc = 6^6.$ In a geometric sequence $ac = b^2,$ hence $b^3 = 6^6,$ so $b = 36$ and $ac = 36^2 = 1296.$

Since the sequence is increasing, $b - a$ is a positive perfect square, so $a = 36 - k^2$ for some $k = 1, \ldots, 5,$ giving candidates $35, 32, 27, 20, 11.$ Also $a$ must divide $1296 = 2^4 \cdot 3^4,$ and of the candidates only $27$ does, with $c = 1296/27 = 48.$

Indeed $27, 36, 48$ is geometric with ratio $\frac{4}{3},$ and $a + b + c = 27 + 36 + 48 = 111.$

2002 AIME II 第 3 题

答案：111

难度评级：2170

解答：

其他年份的第 3 题