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

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== Solution ==
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== Solution 1 ==
 
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>
  

Revision as of 22:36, 9 February 2020

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 1

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 Since there is a $\frac{1}{5}$ chance that sum of digits of a randomly selected positive integer 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 (ProblemsAnswer KeyResources)
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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