1986 AMC 12 第 23 题

Question

设

\begin{aligned} N={}&69^5+5\cdot69^4+10\cdot69^3\\ &{}+10\cdot69^2+5\cdot69+1 \end{aligned}\text{。}

N

有多少个正因数？

Let

\begin{aligned} N={}&69^5+5\cdot69^4+10\cdot69^3\\ &{}+10\cdot69^2+5\cdot69+1. \end{aligned}

How many positive integers are factors of

N?

3

5

69

125

216

识别系数

1,5,10,10,5,1

Recognize the coefficients

1,5,10,10,5,1

使用二项式定理后，将所得的底数分解为质因数

After using the binomial theorem, factor the resulting base into primes

由二项式定理，

\begin{aligned} N&=(69+1)^5=70^5\\ &=2^5\cdot5^5\cdot7^5 \end{aligned}\text{。}

构造一个因数时，可以分别为这三个质因数选择从

0

到

5

的指数。因此，

N

有

6^3=216

个正因数。

因此正确答案是 E。

By the binomial theorem,

\begin{aligned} N&=(69+1)^5=70^5\\ &=2^5\cdot5^5\cdot7^5. \end{aligned}

A divisor independently chooses an exponent from

0

through

5

for each of the three primes. Thus

N

has

6^3=216

positive divisors.

Therefore the correct answer is E.

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