Difference between revisions of "2011 AIME I Problems/Problem 11"
Danielguo94 (talk | contribs) (→Solution) |
(Linked to Euler's theorem) |
||
Line 3: | Line 3: | ||
== 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 be the set of all possible remainders when a number of the form , a nonnegative integer, is divided by . Let be the sum of the elements in . Find the remainder when is divided by .
Solution
Note that and . So we must find the first two integers and such that and and . Note that and will be greater than 2 since remainders of will not be possible after 2 (the numbers following will always be congruent to 0 modulo 8). Note that (see Euler's theorem) and are all distinct modulo 125. Thus, and are the first two integers such that . All that is left is to find in mod . After some computation:
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 |