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

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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}.
 
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To show that <math>2^0, 2^1,\ldots, 2^{99}</math> are distinct modulo 125, suppose for the sake of contradiction that <math>k < 100</math> is the order of 2 modulo 125. Since <math>k|100</math>, we must have at least one of <math>2^{20}\equiv 1\pmod{125}</math> or <math>2^{50}\equiv 1\pmod{125}</math>. However, writing <math>2^{10}\equiv 25 - 1\pmod{125}</math>, we can easily verify that <math>2^{20}\equiv -49\pmod{125}</math> and <math>2^{50}\equiv -1\pmod{125}</math>, giving us the needed contradiction.
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To show that <math>2^0, 2^1,\ldots, 2^{99}</math> are distinct modulo 125, suppose for the sake of contradiction that they are not. Then, we must have at least one of <math>2^{20}\equiv 1\pmod{125}</math> or <math>2^{50}\equiv 1\pmod{125}</math>. However, writing <math>2^{10}\equiv 25 - 1\pmod{125}</math>, we can easily verify that <math>2^{20}\equiv -49\pmod{125}</math> and <math>2^{50}\equiv -1\pmod{125}</math>, giving us the needed contradiction.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2011|n=I|num-b=10|num-a=12}}
 
{{AIME box|year=2011|n=I|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:48, 2 April 2014

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 (proof below). 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: \[S = 2^0+2^1+2^2+2^3+2^4+...+2^{101}+ 2^{102} = 2^{103}-1 \equiv 8 - 1 \mod 1000 = \boxed{007}.\] To show that $2^0, 2^1,\ldots, 2^{99}$ are distinct modulo 125, suppose for the sake of contradiction that they are not. Then, we must have at least one of $2^{20}\equiv 1\pmod{125}$ or $2^{50}\equiv 1\pmod{125}$. However, writing $2^{10}\equiv 25 - 1\pmod{125}$, we can easily verify that $2^{20}\equiv -49\pmod{125}$ and $2^{50}\equiv -1\pmod{125}$, giving us the needed contradiction.

See also

2011 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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