Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
−
Note that the cycle of remainders of <math>2^n</math> will start after <math>2^2</math> because remainders of <math>1, 2, 4</math> 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 <math>2^{100}\equiv 1\mod 125</math>. The order is <math>100</math> starting with remainder <math>8</math>. All that is left is 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 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>S=2^0+2^1+2^2+2^3+2^4...+2^{99}\equiv 2^{100}-1\equiv 8-1\equiv \boxed{007}\mod 1000</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{7}.
 +
</cmath>
  
 
== 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}}

Revision as of 14:18, 30 May 2011

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: \[S = 2^0+2^1+2^2+2^3+2^4+...+2^{101}+ 2^{102} = 2^{103}-1 \equiv 8 - 1 \mod 1000 = \boxed{7}.\]

See also

2011 AIME I (Problems • Answer Key • Resources)
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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_11&oldid=38807"