Difference between revisions of "2012 AIME I Problems/Problem 1"
(→Problem 1) |
|||
| Line 1: | Line 1: | ||
== 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>. | ||
| + | |||
| + | == Solution == | ||
| + | |||
| + | A positive integer is divisible by <math>4</math> if and only if its last two digits are divisible by 4. 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>. Thus, there are <math>2 * 2 = 4</math> ways to choose <math>a</math> and <math>c</math>, and <math>10</math> ways to choose <math>b</math> (since <math>b</math> can be any digit). Therefore, the final answer is <math>4 * 10 = \boxed{040}</math>. | ||
== 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}} | ||
Revision as of 22:19, 16 March 2012
Problem 1
Find the number of positive integers with three not necessarily distinct digits, 
, with 
and 
such that both and 
are multiples of 
.
Solution
A positive integer is divisible by if and only if its last two digits are divisible by 4. 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 
. Thus, there are 
ways to choose and , and 
ways to choose (since can be any digit). Therefore, the final answer is 
.
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 | ||
