Difference between revisions of "2013 AIME I Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | + | The number takes a form of <math>5\text{x,y,z}5</math>, in which <math>5|x+y+z</math>. Let <math>x</math> and <math>y</math> be arbitrary digits. For each pair of <math>x,y</math>, there are exactly two values of <math>z</math> that satisfy the condition of <math>5|x+y+z</math>. Therefore, the answer is <math>10\times10\times2=\boxed{200}</math> | |
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== See also == | == See also == | ||
{{AIME box|year=2013|n=I|num-b=1|num-a=3}} | {{AIME box|year=2013|n=I|num-b=1|num-a=3}} | ||
Revision as of 17:17, 15 March 2013
Problem 2
Find the number of five-digit positive integers, 
, that satisfy the following conditions:
-
(a) the number
is divisible by 
-
(b) the first and last digits of
are equal, and
-
(c) the sum of the digits of
is divisible by 
Solution
The number takes a form of 
, in which 
. Let 
and 
be arbitrary digits. For each pair of 
, there are exactly two values of 
that satisfy the condition of . Therefore, the answer is 
See also
| 2013 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
