You can put this solution on YOUR website! find three-digit numbers
100a + 10b + c
such that its digits are reversed when 99 is added
100a + 10b + c + 99 = 100c + 10b + a
combine like terms on the left
100a - a + 10b - 10b + c - 100c + 99 = 0
99a + 0 - 99c + 99 = 0
simplify, divide by 99
a - c - 1 - 0
a = c + 1
obviously there are alot of numbers like this
102/201, 203/302, 304/403, 435/534, 869/968 are just a few
ANSWER: any number in which the units digit is 1 more than the hundreds digit.
You can see why that is going to work by trying a few examples.
142+99=241
384+99=483
708+99=807
...
We can show that is the answer algebraically:
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.
The original number is
The number when 99 is added is then
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.
