AMC10 2013 B

AMC10 2013 B · Q24

AMC10 2013 B · Q24. It mainly tests Primes & prime factorization, Counting divisors.

A positive integer n is nice if there is a positive integer m with exactly four positive divisors (including 1 and m) such that the sum of the four divisors is equal to n. How many numbers in the set ${2010, 2011, 2012, . . . , 2019}$ are nice?

正整数 n 是“nice”的，如果存在正整数 m 恰有四个正因数（包括 1 和 m），使得四个因数的和等于 n。集合 ${2010, 2011, ..., 2019}$ 中有多少个 nice 数？

(A) 1 1

(B) 2 2

(D) 4 4

(E) 5 5

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Let $n$ denote a nice number from the given set. An integer $m$ has exactly four divisors if and only if $m=p^3$ or $m=pq$, where $p$ and $q$ (with $p>q$) are prime numbers. In the former case, the sum of the four divisor is equal to $1+p+p^2+p^3$. Note that $1+11+11^2+11^3<2010\le n$ and $1+13+13^2+13^3>2019\ge n$. Therefore we must have $m=pq$ and $n=1+q+p+pq=(1+q)(1+p)$. Because $p$ is odd, $n$ must be an even number. If $q=2$, then $n$ must be divisible by $3$. In the given set only $2010=(1+2)(1+669)$ and $2016=(1+2)(1+671)$ satisfy these requirements. However neither $669$ nor $671$ are prime. If $q$ is odd, then $n$ must be divisible by $4$. In the given set, only $2012$ and $2016$ are divisible by $4$. None of the pairs of factors of $2012$, namely $1\cdot 2012$, $2\cdot 1006$, $4\cdot 503$, gives rise to primes $p$ and $q$. This leaves $2016=(1+3)(1+503)$, which is the only nice number in the given set. Remark: Note that $2016$ is nice in five ways. The other four ways are $(1+7)(1+251)$, $(1+11)(1+167)$, $(1+23)(1+83)$, and $(1+41)(1+47)$.

答案（A）：设 $n$ 表示给定集合中的一个“好数”。整数 $m$ 恰有四个正因数当且仅当 $m=p^3$ 或 $m=pq$，其中 $p$ 与 $q$（且 $p>q$）为素数。在前一种情况下，四个因数之和为 $1+p+p^2+p^3$。注意到 $1+11+11^2+11^3<2010\le n$ 且 $1+13+13^2+13^3>2019\ge n$。因此必须有 $m=pq$，并且 $n=1+q+p+pq=(1+q)(1+p)$。因为 $p$ 是奇数，所以 $n$ 必为偶数。若 $q=2$，则 $n$ 必须能被 $3$ 整除。在给定集合中，只有 $2010=(1+2)(1+669)$ 与 $2016=(1+2)(1+671)$ 满足这些要求。然而 $669$ 与 $671$ 都不是素数。若 $q$ 为奇数，则 $n$ 必须能被 $4$ 整除。在给定集合中，只有 $2012$ 与 $2016$ 能被 $4$ 整除。$2012$ 的因数对，即 $1\cdot 2012$、$2\cdot 1006$、$4\cdot 503$，都不能对应得到素数 $p$ 与 $q$。因此只剩下 $2016=(1+3)(1+503)$，它是给定集合中唯一的“好数”。备注：注意 $2016$ 以五种方式是“好”的。另外四种方式为 $(1+7)(1+251)$、$(1+11)(1+167)$、$(1+23)(1+83)$、以及 $(1+41)(1+47)$。

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