AMC12 2010 B

AMC12 2010 B · Q9

AMC12 2010 B · Q9. It mainly tests Primes & prime factorization, Perfect squares & cubes.

Let $n$ be the smallest positive integer such that $n$ is divisible by $20$, $n^2$ is a perfect cube, and $n^3$ is a perfect square. What is the number of digits of $n$?

设$n$为满足以下条件的最小正整数：$n$能被$20$整除，$n^2$是完全立方数，且$n^3$是完全平方数。$n$有多少位数字？

(A) 3 3

(B) 4 4

(D) 6 6

(E) 7 7

Answer

Correct choice: (E)

正确答案：（E）

Solution

We know that $n^2 = k^3$ and $n^3 = m^2$. Cubing and squaring the equalities respectively gives $n^6 = k^9 = m^4$. Let $a = n^6$. Now we know $a$ must be a perfect $36$-th power because $lcm(9,4) = 36$, which means that $n$ must be a perfect $6$-th power. The smallest number whose sixth power is a multiple of $20$ is $10$, because the only prime factors of $20$ are $2$ and $5$, and $10 = 2 \times 5$. Therefore our is equal to number $10^6 = 1000000$, with $7$ digits $\Rightarrow \boxed {E}$.

我们知道$n^2=k^3$且$n^3=m^2$。分别对等式立方与平方得到$n^6=k^9=m^4$。令$a=n^6$。现在$a$必须是完全$36$次幂，因为$lcm(9,4)=36$，这意味着$n$必须是完全$6$次幂。其六次幂是$20$的倍数的最小数是$10$，因为$20$的质因数只有$2$和$5$，且$10=2\times 5$。因此$n=10^6=1000000$，有$7$位数字$\Rightarrow \boxed {E}$。

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