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Difference between revisions of "2013 AIME I Problems/Problem 2"

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== Problem 2 ==
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== 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:
  
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== Solution ==
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== Solution==
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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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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>
  
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== Solution 2 ==
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==Video Solution==
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For a number to be divisible by <math>5</math>, the last digit of the number must be <math>5</math> or <math>0</math>. However, since the first digit is the same as the last one, the last (and first) digits can not be <math>0</math>, so the number must be in the form <math>\overline{5abc5}</math>, where <math>a+b+c</math> is divisible by 5. This gives us a answer of <math>10\times10\times10\times\frac{1}{5}=\boxed{200}</math>
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https://www.youtube.com/watch?v=kz3ZX4PT-_0
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~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}}
 
{{MAA Notice}}

Latest revision as of 18:23, 27 April 2024

Problem

Find the number of five-digit positive integers, $n$, that satisfy the following conditions:

    (a) the number $n$ is divisible by $5,$
    (b) the first and last digits of $n$ are equal, and
    (c) the sum of the digits of $n$ is divisible by $5.$


Solution

The number takes a form of $\overline{5xyz5}$, in which $5|(x+y+z)$. Let $x$ and $y$ be arbitrary digits. For each pair of $x,y$, there are exactly two values of $z$ that satisfy the condition of $5|(x+y+z)$. Therefore, the answer is $10\times10\times2=\boxed{200}$

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

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