Difference between revisions of "2013 AIME I Problems/Problem 2"
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== Solution 1 ==  == Solution 1 ==  
The number takes a form of <math>5\text{x,y,z}5</math>, in which <math>5x+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>5x+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>5x+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>5x+y+z</math>. Therefore, the answer is <math>10\times10\times2=\boxed{200}</math>  
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== See also ==  == See also ==  
{{AIME boxyear=2013n=Inumb=1numa=3}}  {{AIME boxyear=2013n=Inumb=1numa=3}}  
{{MAA Notice}}  {{MAA Notice}} 
Revision as of 14:25, 12 March 2020
Problem 2
Find the number of fivedigit 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 1
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 
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.