# 2007 iTest Problems/Problem 20

## Problem

Find the largest integer $n$ such that $2007^{1024}-1$ is divisible by $2^n$ $\text{(A) } 1\qquad \text{(B) } 2\qquad \text{(C) } 3\qquad \text{(D) } 4\qquad \text{(E) } 5\qquad \text{(F) } 6\qquad \text{(G) } 7\qquad \text{(H) } 8\qquad$ $\text{(I) } 9\qquad \text{(J) } 10\qquad \text{(K) } 11\qquad \text{(L) } 12\qquad \text{(M) } 13\qquad \text{(N) } 14\qquad \text{(O) } 15\qquad \text{(P) } 16\qquad$ $\text{(Q) } 55\qquad \text{(R) } 63\qquad \text{(S) } 64\qquad \text{(T) } 2007\qquad$

## Solution

The expression can be factored by repeatedly using the difference of squares. $$(2007^{512} + 1)(2007^{512} - 1)$$ $$(2007^{512} + 1)(2007^{256} + 1)(2007^{256} - 1)$$ $$(2007^{512} + 1)(2007^{256} + 1) \cdots (2007^1 + 1)(2007^1 - 1)$$ Notice that $2007 \equiv 3 \pmod{4}$, so $2007^2 \equiv 1 \pmod{4}$. Thus, in the expression $2007^a + 1$, if $a$ is even, then the expression is congruent to $2$ modulo $4$.

The remaining numbers to consider are $2008$ and $2006$. Factoring $2008$ yields $8 \cdot 251$, and factoring $2006$ yields $2 \cdot 1003$.

That means $2007^{1004} - 1$ has $9+3+1 = 13$ as the exponent of $2$, so the largest $n$ that makes $2^n$ a factor of $2007^{1004} - 1$ is $\boxed{\text{(M) } 13}$.