AMC12 2019 A

AMC12 2019 A · Q15

AMC12 2019 A · Q15. It mainly tests Logarithms (rare), Word problems (algebra).

Positive real numbers $a$ and $b$ have the property that $\sqrt{\log a} + \sqrt{\log b} + \log \sqrt{a} + \log \sqrt{b} = 100$, and all four terms on the left are positive integers, where $\log$ denotes the base 10 logarithm. What is $ab$?

正实数$a$和$b$满足$\sqrt{\log a} + \sqrt{\log b} + \log \sqrt{a} + \log \sqrt{b} = 100$，且左边四个项均为正整数，其中$\log$表示以10为底的对数。求$ab$？

(A) $10^{52}$ $10^{52}$

(B) $10^{100}$ $10^{100}$

(D) $10^{164}$ $10^{164}$

(E) $10^{200}$ $10^{200}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): If $n$ is a positive integer such that $n=\sqrt{\log a}$, then $a=10^{n^2}$; and if $m$ is a positive integer such that $m=\log\sqrt{a}$, then $a=10^{2m}$. Thus if both expressions are positive integers, then $a=10^{4k^2}$ for some positive integer $k$, and $$ \sqrt{\log a}+\log\sqrt{a}=\sqrt{4k^2}+\log 10^{2k^2}=2k+2k^2. $$ A similar conclusion holds for $b$. The first few possible values of $a$ and $b$ are $10^4$, $10^{16}$, $10^{36}$, $10^{64}$, $10^{100}$, $10^{144}$, and $10^{196}$, and the corresponding values of $\sqrt{\log a}+\log\sqrt{a}$ and $\sqrt{\log b}+\log\sqrt{b}$ less than 100 are 4, 12, 24, 40, 60, and 84. Because the sum of all four terms is 100, it follows that $\{a,b\}=\{10^{64},10^{100}\}$, and $ab=10^{64}\cdot 10^{100}=10^{164}$.

答案（D）：如果$n$是正整数且满足$n=\sqrt{\log a}$，则$a=10^{n^2}$；如果$m$是正整数且满足$m=\log\sqrt{a}$，则$a=10^{2m}$。因此若这两个表达式都是正整数，则对某个正整数$k$有$a=10^{4k^2}$，并且 $$ \sqrt{\log a}+\log\sqrt{a}=\sqrt{4k^2}+\log 10^{2k^2}=2k+2k^2. $$ 对$b$也有类似结论。$a$和$b$的前几个可能取值是$10^4$、$10^{16}$、$10^{36}$、$10^{64}$、$10^{100}$、$10^{144}$和$10^{196}$；对应的$\sqrt{\log a}+\log\sqrt{a}$与$\sqrt{\log b}+\log\sqrt{b}$中小于100的值分别为4、12、24、40、60和84。由于四项之和为100，可得$\{a,b\}=\{10^{64},10^{100}\}$，并且$ab=10^{64}\cdot 10^{100}=10^{164}$。

Topics