2011 AIME I Problems/Problem 11

Revision as of 06:59, 29 March 2011 by Btzy1996 (talk | contribs) (Solution)

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 the cycle of remainders of $2^n$ will start after $2^2$ because remainders of $1, 2, 4$ will not be possible after (the numbers following will always be congruent to 0 modulo 8). Now we have to find the order. Note that $2^{100}\equiv 1\mod 125$. The order is $100$ starting with remainder $8$. All that is left is find $S$ in mod $1000$ after some computation. \[S=2^0+2^1+2^2+2^3+2^4...+2^{102}\equiv 2^{103}-1\equiv 8-1\equiv \boxed{007}\mod 1000\]

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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