Difference between revisions of "2012 AIME I Problems/Problem 1"
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− | == 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>. |
− | + | == 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. | |
− | + | == Video Solutions == | |
+ | |||
+ | https://artofproblemsolving.com/videos/amc/2012aimei/289 | ||
+ | |||
+ | https://youtu.be/ZhAZ1oPe5Ds?t=3235 | ||
+ | |||
+ | https://www.youtube.com/watch?v=T8Ox412AkZc | ||
== See also == | == See also == | ||
{{AIME box|year=2012|n=I|before=First Problem|num-a=2}} | {{AIME box|year=2012|n=I|before=First Problem|num-a=2}} | ||
+ | |||
+ | [[Category:Introductory Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 13:16, 14 February 2021
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.
Video Solutions
https://artofproblemsolving.com/videos/amc/2012aimei/289
https://youtu.be/ZhAZ1oPe5Ds?t=3235
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 |
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