Let's Make All Digits Positive!

Let $A$ and $B$ denote the single distinct digits, such that $45 - B - A = \overline{AB}$ which is equivalent to solving $\begin{array} { l l l } & A & B \\ & & A \\ + & & B \\ \hline & 4 & 5 \\ \end{array}$ Then, $2B + A \equiv 5 \bmod 10$ , which shows that $A$ must be an odd integer. But since the tens digit is even, then the carry digit is $1$ , which forces $A = 3$ and $B = 6$ . Thus, the sum that works is $36$ .

So as long as the possible sum is $36$ , the digits $\{1,2,4,5,7,8,9\}$ are all possible candidates for the equation boxes.

$36=1+2+4+5+7+8+9$

Hence proved.

Easier to solve 45 - a - b = 10a + b => 11a + 2b = 45. Whence a = 3 and b = 6.

The sum of the first 9 digits is 45. And the 2 digits taken away gives an addition of between 3 (1+2) to 17 (9+8). So the sum of the 7 numbers is between 28 and 42. So the maximum digit for the tens column is 3. So the new maximum is 4+9=13. So now the digit in the tens column is either 3 or 4. If it is 4, and an EVEN number, then that must mean that you have taken away an odd number from the 45. So the sum is even - which is a contradiction. And if the tens digit is 4 again, and an ODD number, then that must mean that you have taken away an even number from 45. So the sum is even - another contradiction. So the first digit is 3. 45-3=42 with the sum being thirty-SOMETHING, or to put it another way 30+n. So 30+n=42-n (the right-hand side being the sum without the 3, and then subtracting the n). So n=6. This was a bit of a trial-and-error method, and Michael Huang's method which is the first one is far superior to this one. Regards, David

The two digits not used on the right form the sum on the left.

The sum of 9 digits is 45. The minimum value of 7 digits is 28 = 45 - 8 - 9, the maximum is 43 = 45 - 1 -2.

Thus one of the digits not used on the right must be 2, 3, or 4 to provide a possible leading digit on the left.

With 3 , 45 - 3 = 42 and 42 - 6 = 36 when the two digits not used on the right form the sum on the left.