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Came across an interesting problem was not sure where to begin with it:

The average of two three-digit numbers is found by placing a decimal point between the numbers. Find the sum of the numbers. I know the answer should be $999$. However I am unsure of the steps that lead to this solution.

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  • $\begingroup$ Hint: If the numbers are $a$ and $b$, then $\frac{a+b}{2} = a + \frac{b}{1000}$. $\endgroup$ – Robert Israel Feb 19 '17 at 3:14
  • $\begingroup$ Sorry, can you explain the problem a bit more? $\endgroup$ – Juan Cuiule Feb 19 '17 at 3:19
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Let $x$ and $y$ be two three digit positive integers. If you put a decimal in between $y$ and $x$, like $y.x$ then this is $y+\text{something}$. Because $x$ is a $3$ digit number, like $234$, we have $\text{something}=\frac{x}{1000}$. To see this look at $234$, $\frac{234}{1000}=0.234$.

Without loss of generality the average of the numbers then should be,

$$y+\frac{x}{1000}=\frac{x+y}{2}$$

That is,

$$\frac{x}{500}+2y=x+y$$

Now note $x \in [100,999]$ and $x+y$ must be an integer, and so must $y$. So $\frac{x}{500}$ must be an integer. This is only possible when $x=500$ by our restrictions. In that case we have $1+2y=500+y$ and you can solve for $y$ to get $y=499$. So then the sum is $499+500=999$.

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