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 19:48, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.