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}. | ||
</cmath> | </cmath> | ||
− | To show that <math>2^0, 2^1,\ldots, 2^{99}</math> are distinct modulo 125, suppose for the sake of contradiction that | + | 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 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 (proof below). Thus, and are the first two integers such that . All that is left is to find in mod . After some computation: To show that are distinct modulo 125, suppose for the sake of contradiction that they are not. Then, we must have at least one of or . However, writing , we can easily verify that and , giving us the needed contradiction.
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 |
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