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

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== Solution 2 ==
 
== Solution 2 ==
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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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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> - mathleticguyyy
  
 
== 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}}

Revision as of 20:25, 15 September 2018

Problem 2

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 $5\text{x,y,z}5$, 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 $10\times10\times2=\boxed{200}$

Solution 2

For a number to be divisible by $5$, the last digit of the number must be $5$ or $0$. However, since the first digit is the same as the last one, the last (and first) digits can not be $0$, so the number must be in the form $\overline{5abc5}$, where $a+b+c$ is divisible by 5. This gives us a answer of $10\times10\times10\times\frac{1}{5}=\boxed{200}$ - mathleticguyyy

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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