Difference between revisions of "2012 AIME I Problems/Problem 10"
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== Solution == | == Solution == | ||
− | It is apparent that for a perfect square <math>s^2</math> to satisfy the constraints, we must have <math>s^2 - 256 = 1000n</math> or <math>(s+16)(s-16) = 1000n.</math> Now in order for <math>(s+16)(s-16)</math> to be a multiple of <math>1000,</math> at least one of <math>s+16</math> and <math>s-16</math> must be a multiple of <math>5,</math> and since <math>s+16</math> and <math>s-16</math> are in different residue classes mod <math>5,</math> one term must have all the factors of <math>5</math> and thus must be a multiple of <math>125.</math> Furthermore, each of <math>s+16</math> and <math>s-16</math> must have at least two factors of <math>2,</math> since otherwise <math>(s+16)(s-16)</math> could not possibly be divisible by <math>8.</math> So therefore the conditions are satisfied if either <math>s+16</math> or <math>s-16</math> is divisible by <math>500,</math> or equivalently <math>s = 500n \pm 16.</math> Counting up from <math>n=0</math> to <math>n=5,</math> we see that the tenth value of <math>s</math> is <math>500 \cdot 5 - 16 = 2484</math> and thus <math>t= \frac{2484^2 - 256}{1000} = 6170 \rightarrow \boxed{170.}</math> | + | It is apparent that for a perfect square <math>s^2</math> to satisfy the constraints, we must have <math>s^2 - 256 = 1000n</math> or <math>(s+16)(s-16) = 1000n.</math> Now in order for <math>(s+16)(s-16)</math> to be a multiple of <math>1000,</math> at least one of <math>s+16</math> and <math>s-16</math> must be a multiple of <math>5,</math> and since <math>s+16</math> and <math>s-16</math> are in different residue classes mod <math>5,</math> one term must have all the factors of <math>5</math> and thus must be a multiple of <math>125.</math> Furthermore, each of <math>s+16</math> and <math>s-16</math> must have at least two factors of <math>2,</math> since otherwise <math>(s+16)(s-16)</math> could not possibly be divisible by <math>8.</math> So therefore the conditions are satisfied if either <math>s+16</math> or <math>s-16</math> is divisible by <math>500,</math> or equivalently if <math>s = 500n \pm 16.</math> Counting up from <math>n=0</math> to <math>n=5,</math> we see that the tenth value of <math>s</math> is <math>500 \cdot 5 - 16 = 2484</math> and thus <math>t= \frac{2484^2 - 256}{1000} = 6170 \rightarrow \boxed{170.}</math> |
== See also == | == See also == | ||
{{AIME box|year=2012|n=I|num-b=9|num-a=11}} | {{AIME box|year=2012|n=I|num-b=9|num-a=11}} |
Revision as of 19:20, 17 March 2012
Problem 10
Let be the set of all perfect squares whose rightmost three digits in base
are
. Let
be the set of all numbers of the form
, where
is in
. In other words,
is the set of numbers that result when the last three digits of each number in
are truncated. Find the remainder when the tenth smallest element of
is divided by
.
Solution
It is apparent that for a perfect square to satisfy the constraints, we must have
or
Now in order for
to be a multiple of
at least one of
and
must be a multiple of
and since
and
are in different residue classes mod
one term must have all the factors of
and thus must be a multiple of
Furthermore, each of
and
must have at least two factors of
since otherwise
could not possibly be divisible by
So therefore the conditions are satisfied if either
or
is divisible by
or equivalently if
Counting up from
to
we see that the tenth value of
is
and thus
See also
2012 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |