Difference between revisions of "2012 AIME I Problems/Problem 1"
Mrdavid445 (talk | contribs) |
Mrdavid445 (talk | contribs) |
||
| Line 2: | Line 2: | ||
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>. | ||
| − | ==Solutions | + | ==Solutions== |
=== Solution 1 === | === Solution 1 === | ||
Revision as of 13:38, 24 August 2012
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 | ||
