AMC10 2016 A

AMC10 2016 A · Q22

AMC10 2016 A · Q22. It mainly tests Primes & prime factorization, Counting divisors.

For some positive integer $n$, the number $110n^3$ has 110 positive integer divisors, including 1 and the number $110n^3$. How many positive integer divisors does the number $81n^4$ have?

对于某个正整数 $n$，数 $110n^3$ 有 110 个正整数因数（包括 1 和 $110n^3$ 本身）。那么数 $81n^4$ 有多少个正整数因数？

(A) 110 110

(B) 191 191

(D) 325 325

(E) 425 425

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let $110n^3=p_1^{r_1}p_2^{r_2}\cdots p_k^{r_k}$, where the $p_j$ are distinct primes and the $r_j$ are positive integers. Then $\tau(110n^3)$, the number of positive integer divisors of $110n^3$, is given by $$ \tau(110n^3)=(r_1+1)(r_2+1)\cdots(r_k+1)=110. $$ Because $110=2\cdot5\cdot11$, it follows that $k=3$, $\{p_1,p_2,p_3\}=\{2,5,11\}$, and, without loss of generality, $r_1=1$, $r_2=4$, and $r_3=10$. Therefore $$ n^3=\frac{p_1\cdot p_2^4\cdot p_3^{10}}{110}=p_2^3\cdot p_3^9,\ \text{so}\ n=p_2\cdot p_3^3. $$ It follows that $81n^4=3^4\cdot p_2^4\cdot p_3^{12}$, and because $3$, $p_2$, and $p_3$ are distinct primes, $$ \tau(81n^4)=5\cdot5\cdot13=325. $$

答案（D）：设 $110n^3=p_1^{r_1}p_2^{r_2}\cdots p_k^{r_k}$，其中 $p_j$ 是互不相同的素数，$r_j$ 是正整数。则 $\tau(110n^3)$（即 $110n^3$ 的正因数个数）为 $$ \tau(110n^3)=(r_1+1)(r_2+1)\cdots(r_k+1)=110. $$ 因为 $110=2\cdot5\cdot11$，所以 $k=3$，$\{p_1,p_2,p_3\}=\{2,5,11\}$，并且不失一般性地取 $r_1=1,\ r_2=4,\ r_3=10$。因此 $$ n^3=\frac{p_1\cdot p_2^4\cdot p_3^{10}}{110}=p_2^3\cdot p_3^9,\ \text{所以}\ n=p_2\cdot p_3^3. $$ 从而 $81n^4=3^4\cdot p_2^4\cdot p_3^{12}$。又因为 $3$、$p_2$、$p_3$ 是互不相同的素数， $$ \tau(81n^4)=5\cdot5\cdot13=325. $$

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