Difference between revisions of "2007 iTest Problems/Problem 20"

Revision as of 20:33, 4 February 2023

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 $\boxed{\text{(M) } 13}$ .

2007 iTest (Problems, Answer Key)
Preceded by: Problem 19	Followed by: Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • TB1 • TB2 • TB3 • TB4

@@ Line 19: / Line 19: @@
 \text{(O) } 15\qquad
 \text{(P) } 16\qquad</math>
 <math>\text{(Q) } 55\qquad
 \text{(R) } 63\qquad

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Difference between revisions of "2007 iTest Problems/Problem 20"

Revision as of 20:33, 4 February 2023

Problem

Solution

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