1986 AHSME Problems/Problem 23

Problem

Let N = $69^{5} + 5\cdot69^{4} + 10\cdot69^{3} + 10\cdot69^{2} + 5\cdot69 + 1$ . 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 $69=a$ . 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 the desired sum is equal to $70^5$ which can be prime factorized as $2^5\cdot5^5\cdot7^5$ . Finally, we can count the number of factors of this number.

$6\cdot6\cdot6=\fbox{(E) 216}$ .

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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