Difference between revisions of "2012 AIME I Problems/Problem 1"
Pi is 3.14 (talk | contribs) (→Video Solution) |
Pi is 3.14 (talk | contribs) (→Video Solutions) |
||
(5 intermediate revisions by 3 users not shown) | |||
Line 1: | Line 1: | ||
− | == Problem | + | == Problem== |
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>. | ||
− | + | == Solution 1 == | |
− | |||
− | |||
A positive integer is divisible by <math>4</math> if and only if its last two digits are divisible by <math>4.</math> For any value of <math>b</math>, there are two possible values for <math>a</math> and <math>c</math>, since we find that if <math>b</math> is even, <math>a</math> and <math>c</math> must be either <math>4</math> or <math>8</math>, and if <math>b</math> is odd, <math>a</math> and <math>c</math> must be either <math>2</math> or <math>6</math>. There are thus <math>2 \cdot 2 = 4</math> ways to choose <math>a</math> and <math>c</math> for each <math>b,</math> and <math>10</math> ways to choose <math>b</math> since <math>b</math> can be any digit. The final answer is then <math>4 \cdot 10 = \boxed{040}</math>. | A positive integer is divisible by <math>4</math> if and only if its last two digits are divisible by <math>4.</math> For any value of <math>b</math>, there are two possible values for <math>a</math> and <math>c</math>, since we find that if <math>b</math> is even, <math>a</math> and <math>c</math> must be either <math>4</math> or <math>8</math>, and if <math>b</math> is odd, <math>a</math> and <math>c</math> must be either <math>2</math> or <math>6</math>. There are thus <math>2 \cdot 2 = 4</math> ways to choose <math>a</math> and <math>c</math> for each <math>b,</math> and <math>10</math> ways to choose <math>b</math> since <math>b</math> can be any digit. The final answer is then <math>4 \cdot 10 = \boxed{040}</math>. | ||
− | + | == Solution 2 == | |
A number is divisible by four if its last two digits are divisible by 4. Thus, we require that <math> 10b + a </math> and <math> 10b + c</math> are both divisible by <math> 4 </math>. If <math> b </math> is odd, then <math> a </math> and <math> c </math> must both be <math> 2 \pmod 4 </math> meaning that <math> a </math> and <math> c </math> are <math> 2 </math> or <math> 6 </math>. If <math> b </math> is even, then <math> a </math> and <math> c </math> must be <math> 0 \pmod 4 </math> meaning that <math> a </math> and <math> c </math> are <math> 4 </math> or <math> 8 </math>. For each choice of <math> b </math> there are <math> 2 </math> choices for <math> a </math> and <math> 2 </math> for <math> c </math> for a total of <math> 10 \cdot 2 \cdot 2 = \boxed{040} </math> numbers. | A number is divisible by four if its last two digits are divisible by 4. Thus, we require that <math> 10b + a </math> and <math> 10b + c</math> are both divisible by <math> 4 </math>. If <math> b </math> is odd, then <math> a </math> and <math> c </math> must both be <math> 2 \pmod 4 </math> meaning that <math> a </math> and <math> c </math> are <math> 2 </math> or <math> 6 </math>. If <math> b </math> is even, then <math> a </math> and <math> c </math> must be <math> 0 \pmod 4 </math> meaning that <math> a </math> and <math> c </math> are <math> 4 </math> or <math> 8 </math>. For each choice of <math> b </math> there are <math> 2 </math> choices for <math> a </math> and <math> 2 </math> for <math> c </math> for a total of <math> 10 \cdot 2 \cdot 2 = \boxed{040} </math> numbers. | ||
− | == | + | == Solution 3 == |
− | + | For this number to fit the requirements <math>bc</math> and <math>ba</math> must be divisible by 4. So <math>bc = 00, 04, 08, 12, 16, ... , 92, 96</math> and so must <math>ba</math> for each two digits of <math>bc</math>. There are two possibilities for <math>ba</math> if <math>b</math> is odd and three possibilities if <math>b</math> is even. So there are <math>2^{2} \cdot 5+3^{2} \cdot 4 = 65</math> possibilities but this overcounts when <math>a</math> or <math>c = 0</math>. So when <math>bc = 00, 20, 40, 60, 80</math> and the corresponding <math>ba</math> should be removed, so <math>65 - 5 \cdot 3 = 50</math>. But we are still overcounting when <math>b</math> is even because then <math>a</math> can be 0. So the answer is <math>50 - 10 \cdot 1 = \boxed{040}</math> | |
− | ~ | + | ~LuisFonseca123 |
− | == Video Solution == | + | == Video Solution by OmegaLearn == |
https://youtu.be/ZhAZ1oPe5Ds?t=3235 | https://youtu.be/ZhAZ1oPe5Ds?t=3235 | ||
~ pi_is_3.14 | ~ pi_is_3.14 | ||
− | ==Video | + | == Video Solutions == |
+ | |||
+ | https://artofproblemsolving.com/videos/amc/2012aimei/289 | ||
https://www.youtube.com/watch?v=T8Ox412AkZc | https://www.youtube.com/watch?v=T8Ox412AkZc | ||
− | |||
== See also == | == See also == |
Latest revision as of 03:59, 21 January 2023
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 | |
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.