AMC12 2000 A

AMC12 2000 A · Q23

AMC12 2000 A · Q23. It mainly tests Basic counting (rules of product/sum), Probability (basic).

Professor Gamble buys a lottery ticket, which requires that he pick six different integers from $1$ through $46$, inclusive. He chooses his numbers so that the sum of the base-ten logarithms of his six numbers is an integer. It so happens that the integers on the winning ticket have the same property— the sum of the base-ten logarithms is an integer. What is the probability that Professor Gamble holds the winning ticket?

甘布尔教授买了一张彩票，要求他从 $1$ 到 $46$（含）中选出六个不同的整数。他选择这些数使得这六个数的以 10 为底的对数之和为整数。碰巧，中奖彩票上的整数也具有同样的性质——以 10 为底的对数之和为整数。甘布尔教授持有中奖彩票的概率是多少？

(A) 1/5 1/5

(B) 1/4 1/4

(D) 1/2 1/2

(E) 1 1

Answer

Correct choice: (B)

正确答案：（B）

Solution

The product of the numbers has to be a power of $10$ in order to have an integer base ten logarithm. Thus all of the numbers must be in the form $2^m5^n$. Listing out such numbers from $1$ to $46$, we find $1,2,4,5,8,10,16,20,25,32,40$ are the only such numbers. Immediately it should be noticed that there are a larger number of powers of $2$ than of $5$. Since a number in the form of $10^k$ must have the same number of $2$s and $5$s in its factorization, we require larger powers of $5$ than we do of $2$. To see this, for each number subtract the power of $5$ from the power of $2$. This yields $0,1,2,-1,3,0,4,1,-2,5,2$, and indeed the only non-positive terms are $0,0,-1,-2$. Since there are only two zeros, the largest number that Professor Gamble could have picked would be $2$. Thus Gamble picks numbers which fit $-2 + -1 + 0 + 0 + 1 + 2$, with the first four having already been determined to be $\{25,5,1,10\}$. The choices for the $1$ include $\{2,20\}$ and the choices for the $2$ include $\{4,40\}$. Together these give four possible tickets, which makes Professor Gamble’s probability $1/4\ \mathrm{(B)}$.

要使以 10 为底的对数之和为整数，这些数的乘积必须是 $10$ 的幂。因此所有数都必须形如 $2^m5^n$。列出 $1$ 到 $46$ 中这样的数，得到 $1,2,4,5,8,10,16,20,25,32,40$ 是仅有的这些数。立刻注意到 $2$ 的幂比 $5$ 的幂多。由于形如 $10^k$ 的数在分解中必须含有相同个数的 $2$ 和 $5$，我们需要比 $2$ 更大的 $5$ 的幂。为此，对每个数用“$2$ 的指数减去 $5$ 的指数”，得到 $0,1,2,-1,3,0,4,1,-2,5,2$，确实只有非正项为 $0,0,-1,-2$。由于只有两个 $0$，甘布尔教授能选到的最大数只能是 $2$。因此甘布尔教授选的数必须对应于 $-2 + -1 + 0 + 0 + 1 + 2$，其中前四个已确定为 $\{25,5,1,10\}$。对应 $1$ 的选择有 $\{2,20\}$，对应 $2$ 的选择有 $\{4,40\}$。合起来共有四种可能的彩票，因此甘布尔教授中奖的概率为 $1/4\ \mathrm{(B)}$。

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