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− | ==Problem==
| + | What is 1+1? |
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− | Find the largest integer <math>n</math> such that <math>2007^{1024}-1</math> is divisible by <math>2^n</math>
| + | No answer for the next few years, try again later! Thanks for your patience. |
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− | <math>\text{(A) } 1\qquad
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− | \text{(B) } 2\qquad
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− | \text{(C) } 3\qquad
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− | \text{(D) } 4\qquad
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− | \text{(E) } 5\qquad
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− | \text{(F) } 6\qquad
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− | \text{(G) } 7\qquad
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− | \text{(H) } 8\qquad</math>
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− | <math>\text{(I) } 9\qquad
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− | \text{(J) } 10\qquad
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− | \text{(K) } 11\qquad
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− | \text{(L) } 12\qquad
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− | \text{(M) } 13\qquad
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− | \text{(N) } 14\qquad
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− | \text{(O) } 15\qquad
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− | \text{(P) } 16\qquad</math>
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− | <math>\text{(Q) } 55\qquad
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− | \text{(R) } 63\qquad
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− | \text{(S) } 64\qquad
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− | \text{(T) } 2007\qquad</math>
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− | ==Solution==
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− | The expression can be factored by repeatedly using the difference of squares.
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− | <cmath>(2007^{512} + 1)(2007^{512} - 1)</cmath>
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− | <cmath>(2007^{512} + 1)(2007^{256} + 1)(2007^{256} - 1)</cmath>
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− | <cmath>(2007^{512} + 1)(2007^{256} + 1) \cdots (2007^1 + 1)(2007^1 - 1)</cmath>
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− | Notice that <math>2007 \equiv 3 \pmod{4}</math>, so <math>2007^2 \equiv 1 \pmod{4}</math>. Thus, in the expression <math>2007^a + 1</math>, if <math>a</math> is even, then the expression is congruent to <math>2</math> [[modulo]] <math>4</math>.
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− | <br>
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− | The remaining numbers to consider are <math>2008</math> and <math>2006</math>. Factoring <math>2008</math> yields <math>8 \cdot 251</math>, and factoring <math>2006</math> yields <math>2 \cdot 1003</math>.
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− | <br>
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− | That means <math>2007^{1004} - 1</math> has <math>9+3+1 = 13</math> as the exponent of <math>2</math>, so the largest <math>n</math> that makes <math>2^n</math> a factor of <math>2007^{1004} - 1</math> is <math>\boxed{\text{(M) } 13}</math>.
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| ==See Also== | | ==See Also== |
Latest revision as of 22:03, 8 May 2024
What is 1+1?
No answer for the next few years, try again later! Thanks for your patience.
See Also