AMC10 2019 A

AMC10 2019 A · Q11

AMC10 2019 A · Q11. It mainly tests Primes & prime factorization, Perfect squares & cubes.

How many positive integer divisors of $201^9$ are perfect squares or perfect cubes (or both)?

$201^9$ 有多少个正整数除数是完全平方数或完全立方数（或两者皆是）？

(A) 32 32

(B) 36 36

(D) 39 39

(E) 41 41

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Because $201^9 = 3^9 \cdot 67^9$, a square divisor has the form $3^a \cdot 67^b$ where $a, b \in \{0,2,4,6,8\}$, and a cubic divisor has the form $3^a \cdot 67^b$ where $a, b \in \{0,3,6,9\}$. A number is both a square and a cube if and only if it is a sixth power, so it has the form $3^a \cdot 67^b$ where $a, b \in \{0,6\}$. Thus there are $5 \cdot 5 = 25$ square divisors, $4 \cdot 4 = 16$ cubic divisors, and $2 \cdot 2 = 4$ divisors that are sixth powers. Therefore the number of divisors that are squares and/or cubes is $25 + 16 - 4 = 37$.

答案（C）：因为 $201^9 = 3^9 \cdot 67^9$，一个平方因数可写成 $3^a \cdot 67^b$，其中 $a, b \in \{0,2,4,6,8\}$；一个立方因数可写成 $3^a \cdot 67^b$，其中 $a, b \in \{0,3,6,9\}$。一个数既是平方又是立方，当且仅当它是六次幂，因此它可写成 $3^a \cdot 67^b$，其中 $a, b \in \{0,6\}$。所以平方因数有 $5 \cdot 5 = 25$ 个，立方因数有 $4 \cdot 4 = 16$ 个，六次幂因数有 $2 \cdot 2 = 4$ 个。因此既是平方和/或立方的因数个数为 $25 + 16 - 4 = 37$。

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