Question 1193692
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ANSWER: any number in which the units digit is 1 more than the hundreds digit.<br>
You can see why that is going to work by trying a few examples.<br>
142+99=241
384+99=483
708+99=807
...<br>
We can show that is the answer algebraically:<br>
Let a be the hundreds digit; then a+1 is the units digit.  We can call the tens digit b; but it will turn out that b can be any digit.<br>
The original number is<br>
{{{100(a)+10(b)+1(a+1) = 100a+10b+a+1 = 101a+10b+1}}}<br>
The number when 99 is added is then<br>
{{{101a+10b+1+99 = 101a+10b+100 = 100a+100+10b+a = 100(a+1)+10b+a}}}<br>
In that number, the hundreds digit is a+1, which was the units digit of the original number; the tens digit remains the same at b; and the units digit is a, which was the hundreds digit of the original number.  So the number after adding 99 to the original number is the original number with the digits reversed.<br>