1986 AHSME Problems/Problem 23

Problem

Let N = $69^{5} + 5*69^{4} + 10*69^{3} + 10*69^{2} + 5*69 + 1$ . How many positive integers are factors of $N$ ?

$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 69\qquad \textbf{(D)}\ 125\qquad \textbf{(E)}\ 216$

Solution

Let 69 be a. Therefore, the equation becomes a^5+5a^4+10a^3+10a^2+5a+1. From Pascal's Triangle, we know this equation is equal to (a+1)^5 . Simplifying, we have 70^5 which can be prime factorized to 2^5x5^5x7^5. Finally, we can figure out how many factors this has .

6x6x6=216.

1986 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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1986 AHSME Problems/Problem 23

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