In the addition of two 2-digit numbers, each blank space, including those in the answer, is to be filled with one of the digits $0,1,2,3,4,5,6$, each used exactly once. The units digit of the sum is
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-073.jpg?height=260&width=311&top_left_y=393&top_left_x=1292)

Part C: Each correct answer is worth 8.
We label the blank spaces to make them easier to refer to.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-327.jpg?height=249&width=309&top_left_y=1385&top_left_x=970)

Since we are adding two 2-digit numbers, then their sum cannot be 200 or greater, so $E$ must be a 1 if the sum is to have 3 digits.

Where can the digit 0 go?

Since no number can begin with a 0 , then neither $A$ nor $C$ can be 0 .

Since each digit is different, then neither $B$ and $D$ can be 0 , otherwise both $D$ and $G$ or $B$ and $G$ would be the same.

Therefore, only $F$ or $G$ could be 0 .

Since we are adding two 2-digit numbers and getting a number which is at least 100, then $A+C$ must be at least 9 . (It could be 9 if there was a "carry" from the sum of the units digits.) This tells us that $A$ and $C$ must be 3 and 6,4 and 5, 4 and 6 , or 5 and 6 .

If $G$ was 0 , then $B$ and $D$ would have to 4 and 6 in some order. But then the largest that $A$ and $C$ could be would be 3 and 5 , which are not among the possibilities above. Therefore, $G$ is not 0 , so $F=0$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-328.jpg?height=249&width=298&top_left_y=152&top_left_x=968)

So the sum of $A$ and $C$ is either 9 or 10 , so $A$ and $C$ are 3 and 6,4 and 5 , or 4 and 6 . In any of these cases, the remaining possibilities for $B$ and $D$ are too small to give a carry from the units column to the tens column.

So in fact, $A$ and $C$ must add to 10 , so $A$ and $C$ are 4 and 6 in some order.

Let's try $A=4$ and $C=6$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-328.jpg?height=252&width=295&top_left_y=725&top_left_x=969)

The remaining digits are 2, 3 and 5 . To make the addition work, $B$ and $D$ must be 2 and 3 and $G$ must be 5 . (We can check that either order for $B$ and $D$ works, and that switching the 4 and 6 will also work.)

So the units digit of the sum must be 5 , as in the example

|  | 4 | 2 |
| :---: | :---: | :---: |
| + | 6 | 3 |
| 1 | 0 | 5 |

(Note that we could have come up with this answer by trial and error instead of this logical procedure.)

ANswer: (D)