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

m (→‎Problem)
(Solution to Problem 20)
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== Problem ==
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==Problem==
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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>
 
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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\text{(T) } 2007\qquad</math>
 
\text{(T) } 2007\qquad</math>
  
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== Solution ==
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==Solution==
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The expression can be factored by repeatedly using the difference of squares.
 +
<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==
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{{iTest box|year=2007|num-b=19|num-a=21}}
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[[Category:Intermediate Number Theory Problems]]

Revision as of 20:23, 28 July 2018

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

See Also

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