Difference between revisions of "2011 AIME I Problems/Problem 11"
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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 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{ | + | 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}. |
</cmath> | </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 00:51, 16 October 2011
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 |