AMC12 2020 A

AMC12 2020 A · Q13

AMC12 2020 A · Q13. It mainly tests Exponents & radicals, Number theory misc.

There are integers $a$, $b$, and $c$, each greater than 1, such that $\sqrt[a]{N}\sqrt[b]{N}\sqrt[c]{N}=\sqrt[36]{N^{25}}$ for all $N>1$. What is $b$?

存在整数 $a$、$b$ 和 $c$，每个都大于 1，使得 $\sqrt[a]{N}\sqrt[b]{N}\sqrt[c]{N}=\sqrt[36]{N^{25}}$ 对所有 $N>1$ 成立。$b$ 是多少？

(A) 2 2

(B) 3 3

(D) 5 5

(E) 6 6

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Written with exponential notation, the left-hand side of the given equation is $((N^{\frac{1}{c}}\cdot N^1)^{\frac{1}{b}}\cdot N^1)^{\frac{1}{a}}=((N^{\frac{1+c}{c}})^{\frac{1}{b}}\cdot N^1)^{\frac{1}{a}}$ $=(N^{\frac{1+c+bc}{bc}})^{\frac{1}{a}}$ $=N^{\frac{1+c+bc}{abc}}.$ Equating exponents and multiplying by $a$ gives $\frac{1+c+bc}{bc}=a\cdot\frac{25}{36}.$ Because $b$ and $c$ are given to be greater than 1, the left-hand side is less than 2, so $a$ cannot be as large as 3. Because $a>1$ as well, $a$ must equal 2, and this equation is equivalent to $18(bc+c+1)=25bc$, from which it follows that $18(c+1)=7bc$. Therefore $c$ divides 18 and 7 divides $c+1$, and thus $c=6$ and $b=3$.

答案（B）：用指数表示法写出，所给等式的左边为 $((N^{\frac{1}{c}}\cdot N^1)^{\frac{1}{b}}\cdot N^1)^{\frac{1}{a}}=((N^{\frac{1+c}{c}})^{\frac{1}{b}}\cdot N^1)^{\frac{1}{a}}$ $=(N^{\frac{1+c+bc}{bc}})^{\frac{1}{a}}$ $=N^{\frac{1+c+bc}{abc}}.$ 令指数相等并两边乘以 $a$，得到 $\frac{1+c+bc}{bc}=a\cdot\frac{25}{36}.$ 因为已知 $b$ 和 $c$ 都大于 1，所以左边小于 2，因此 $a$ 不可能达到 3。又因为 $a>1$，所以 $a$ 必须等于 2。该方程等价于 $18(bc+c+1)=25bc$，从而推出 $18(c+1)=7bc$。因此 $c$ 是 18 的因数且 7 整除 $c+1$，于是 $c=6$ 且 $b=3$。

Topics