Difference between revisions of "2012 AIME I Problems/Problem 10"
m (→Solution) |
|||
| Line 8: | Line 8: | ||
== 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}} | ||
| + | {{MAA Notice}} | ||
Revision as of 19:28, 4 July 2013
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 the corresponding element in is 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
