Let N=&69^5+5×69^4+10×69^3 &+10×69^2+5×69+1. How many positive integers are factors of N?

1986 AMC 12 Problem 23

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23.

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

Recognize the coefficients

1,5,10,10,5,1

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

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.