Difference between revisions of "2012 AIME I Problems/Problem 1"
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== Problem 1 == | == Problem 1 == | ||
Find the number of positive integers with three not necessarily distinct digits, <math>abc</math>, with <math>a \neq 0</math> and <math>c \neq 0</math> such that both <math>abc</math> and <math>cba</math> are multiples of <math>4</math>. | Find the number of positive integers with three not necessarily distinct digits, <math>abc</math>, with <math>a \neq 0</math> and <math>c \neq 0</math> such that both <math>abc</math> and <math>cba</math> are multiples of <math>4</math>. |
Revision as of 15:42, 9 August 2018
Problem
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.
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 |
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