Replace the letters in this equation by digits such that it becomes a sound mathematical expression. The same letters are the same digits and different letters are different digits. No number with more than one digit has leading zeros.


To solve the example puzzle let's first think about what the A can be, occuring at the start of the largest number. An upper bound for the sum of the two smaller numbers would be 99+99=198. Of course that upper bound is too high since not all of the digits A, B and C can be 9, and with a little more thought we could put it at 97+89=186, but 198 is good enough to conclude that A is at most 1, and since it can not be a (leading) zero, it must be exactly 1. Also the other A's can be replaced by a 1.

Next we will concentrate on the value of D. With A being 1 and C at most 9, A+C can be at most 10, but we also have a potential carry of 1 from A+B in case it is larger than 9. AD however can not represent 11, and it must be at least 10, so it is exactly 10 and D is zero.

Now C can be 9 in case A+B does not give a carry, or 8 if it does. Suppose that it does and C is in fact 8, then A+B should be 18 and this is impossible. We can conclude C is 9 and consequently B must be 8.

Puzzles in this genre.