Difference between revisions of "2013 AIME I Problems/Problem 2"
(→Solution) |
Mathkiddie (talk | contribs) (→Solution) |
||
Line 16: | Line 16: | ||
== Solution== | == Solution== | ||
− | The number takes a form of <math> | + | 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== | ==Video Solution== |
Revision as of 18:23, 27 April 2024
Contents
[hide]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.