2012 AIME I Problems/Problem 1
Contents
Problem
Find the number of positive integers with three not necessarily distinct digits, , with
and
such that both
and
are multiples of
.
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.
Solution 3
For this number to fit the requirements and
must be divisible by 4. So
and so must
for each two digits of
. There are two possibilities for
if
is odd and three possibilities if
is even. So there are
possibilities but this overcounts when
or
. So when
and the corresponding
should be removed, so
. But we are still overcounting when
is even because then
can be 0. So the answer is
~LuisFonseca123
Video Solution by OmegaLearn
https://youtu.be/ZhAZ1oPe5Ds?t=3235
~ pi_is_3.14
Video Solutions
https://artofproblemsolving.com/videos/amc/2012aimei/289
https://www.youtube.com/watch?v=T8Ox412AkZc
See also
2012 AIME I (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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