Difference between revisions of "2011 AIME I Problems/Problem 11"

(Solution)
(Linked to Euler's theorem)
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== Solution ==
 
== Solution ==
Note that <math>x \equiv y \pmod{1000} \Leftrightarrow x \equiv y \pmod{125}</math> and <math>x \equiv y \pmod{8}</math>. So we must find the first two integers <math>i</math> and <math>j</math> such that <math>2^i \equiv 2^j \pmod{125}</math> and <math>2^i \equiv 2^j \pmod{8}</math> and <math>i \neq j</math>. Note that <math>i</math> and <math>j</math> will be greater than 2 since remainders of <math>1, 2, 4</math> will not be possible after 2 (the numbers following will always be congruent to 0 modulo 8). Note that <math>2^{100}\equiv 1\pmod{125}</math> (see Euler's theorem) and <math>2^0,2^1,2^2,\ldots,2^{99}</math> are all distinct modulo 125. Thus, <math>i = 3</math> and <math>j =103</math> are the first two integers such that <math>2^i \equiv 2^j \pmod{1000}</math>. All that is left is to find <math>S</math> in mod <math>1000</math>. After some computation:
+
Note that <math>x \equiv y \pmod{1000} \Leftrightarrow x \equiv y \pmod{125}</math> and <math>x \equiv y \pmod{8}</math>. So we must find the first two integers <math>i</math> and <math>j</math> such that <math>2^i \equiv 2^j \pmod{125}</math> and <math>2^i \equiv 2^j \pmod{8}</math> and <math>i \neq j</math>. Note that <math>i</math> and <math>j</math> will be greater than 2 since remainders of <math>1, 2, 4</math> will not be possible after 2 (the numbers following will always be congruent to 0 modulo 8). Note that <math>2^{100}\equiv 1\pmod{125}</math> (see [[Euler's Totient Theorem|Euler's theorem]]) and <math>2^0,2^1,2^2,\ldots,2^{99}</math> are all distinct modulo 125. Thus, <math>i = 3</math> and <math>j =103</math> are the first two integers such that <math>2^i \equiv 2^j \pmod{1000}</math>. All that is left is to find <math>S</math> in mod <math>1000</math>. After some computation:
 
<cmath>
 
<cmath>
 
S = 2^0+2^1+2^2+2^3+2^4+...+2^{101}+ 2^{102} = 2^{103}-1 \equiv 8 - 1 \mod 1000 = \boxed{007}.
 
S = 2^0+2^1+2^2+2^3+2^4+...+2^{101}+ 2^{102} = 2^{103}-1 \equiv 8 - 1 \mod 1000 = \boxed{007}.

Revision as of 12:35, 31 March 2013

Problem

Let $R$ be the set of all possible remainders when a number of the form $2^n$, $n$ a nonnegative integer, is divided by $1000$. Let $S$ be the sum of the elements in $R$. Find the remainder when $S$ is divided by $1000$.

Solution

Note that $x \equiv y \pmod{1000} \Leftrightarrow x \equiv y \pmod{125}$ and $x \equiv y \pmod{8}$. So we must find the first two integers $i$ and $j$ such that $2^i \equiv 2^j \pmod{125}$ and $2^i \equiv 2^j \pmod{8}$ and $i \neq j$. Note that $i$ and $j$ will be greater than 2 since remainders of $1, 2, 4$ will not be possible after 2 (the numbers following will always be congruent to 0 modulo 8). Note that $2^{100}\equiv 1\pmod{125}$ (see Euler's theorem) and $2^0,2^1,2^2,\ldots,2^{99}$ are all distinct modulo 125. Thus, $i = 3$ and $j =103$ are the first two integers such that $2^i \equiv 2^j \pmod{1000}$. All that is left is to find $S$ in mod $1000$. After some computation: \[S = 2^0+2^1+2^2+2^3+2^4+...+2^{101}+ 2^{102} = 2^{103}-1 \equiv 8 - 1 \mod 1000 = \boxed{007}.\]

See also

2011 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AIME Problems and Solutions
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