Difference between revisions of "2013 AIME I Problems/Problem 2"
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| − | == Problem | + | == Problem == |
Find the number of five-digit positive integers, <math>n</math>, that satisfy the following conditions: | Find the number of five-digit positive integers, <math>n</math>, that satisfy the following conditions: | ||
| Line 15: | Line 15: | ||
| − | == Solution == | + | == Solution== |
| + | The number takes a form of <math>\overline{5xyz5}</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> | ||
| − | + | ==Video Solution== | |
| + | |||
| + | https://www.youtube.com/watch?v=kz3ZX4PT-_0 | ||
| + | ~Shreyas S | ||
== 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}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:23, 27 April 2024
Contents
Problem
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 
Video Solution
https://www.youtube.com/watch?v=kz3ZX4PT-_0 ~Shreyas S
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
