Difference between revisions of "2012 AIME I Problems/Problem 1"
m (→Video Solution by Richard Rusczyk) |
|||
| Line 10: | Line 10: | ||
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. | ||
| − | == Video | + | == Video Solutions == |
https://artofproblemsolving.com/videos/amc/2012aimei/289 | https://artofproblemsolving.com/videos/amc/2012aimei/289 | ||
| − | + | https://youtu.be/ZhAZ1oPe5Ds?t=3235 | |
| + | |||
| + | https://www.youtube.com/watch?v=T8Ox412AkZc | ||
== Video Solution == | == Video Solution == | ||
Revision as of 14:15, 14 February 2021
Contents
[hide]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
Video Solution
https://youtu.be/ZhAZ1oPe5Ds?t=3235
~ pi_is_3.14
Video Solution
https://www.youtube.com/watch?v=T8Ox412AkZc ~Shreyas S
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. 
