Difference between revisions of "2012 AIME I Problems/Problem 1"
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== See also == | == See also == |
Revision as of 12:32, 15 May 2020
Contents
Problem 1
Find the number of positive integers with three not necessarily distinct digits, , with
and
such that both
and
are multiples of
.
Solutions
Solution 1
A positive integer is divisible by if and only if its last two digits are divisible by
For any value of
, there are two possible values for
and
, since we find that if
is even,
and
must be either
or
, and if
is odd,
and
must be either
or
. There are thus
ways to choose
and
for each
and
ways to choose
since
can be any digit. The final answer is then
.
Solution 2
A number is divisible by four if its last two digits are divisible by 4. Thus, we require that and
are both divisible by
. If
is odd, then
and
must both be
meaning that
and
are
or
. If
is even, then
and
must be
meaning that
and
are
or
. For each choice of
there are
choices for
and
for
for a total of
numbers.
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012aimei/289
~ dolphin7
See also
2012 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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.