A Big Number

Number Theory Level pending

Let $A_k = 1000...128$ where there are $k$ zeroes separating $1$ from $128$ . Let $N(k)$ be the number of factors of two in the prime factorization of $A_k$ . Find the maximum value of $N(k)$ .

8 7 5 6

1 solution

Alan Yan
Sep 14, 2015

$A_k = 10^{k+3} + 2^7 = 2^7(2^{k-4}\cdot 5^{k+3} + 1)$

You may be inclined to stop now and choose $7$ as your answer, however, notice if we substitute $k = 4$ , we get another factor of two.

$k = 4 \text{ | } 2^7(5^7 + 1) = 2^8(1563)$

Therefore there are $\boxed{8}$ factors.

A Big Number

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