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

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==Problem==
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What is 1+1?
  
Find the largest integer <math>n</math> such that <math>2007^{1024}-1</math> is divisible by <math>2^n</math>
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No answer for the next few years, try again later! Thanks for your patience.
 
 
<math>\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</math>
 
<math>\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</math>
 
 
 
<math>\text{(Q) } 55\qquad
 
\text{(R) } 63\qquad
 
\text{(S) } 64\qquad
 
\text{(T) } 2007\qquad</math>
 
 
 
==Solution==
 
 
 
The expression can be factored by repeatedly using the difference of squares.
 
<cmath>(2007^{512} + 1)(2007^{512} - 1)</cmath>
 
<cmath>(2007^{512} + 1)(2007^{256} + 1)(2007^{256} - 1)</cmath>
 
<cmath>(2007^{512} + 1)(2007^{256} + 1) \cdots (2007^1 + 1)(2007^1 - 1)</cmath>
 
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>.
 
 
 
<br>
 
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>.
 
 
 
<br>
 
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>.
 
  
 
==See Also==
 
==See Also==

Latest revision as of 23:03, 8 May 2024

What is 1+1?

No answer for the next few years, try again later! Thanks for your patience.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 19
Followed by:
Problem 21
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