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> | 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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==Video Solution== | ==Video Solution== | ||
Revision as of 01:30, 3 October 2020
Contents
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 
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. 
