Two different digits from 1 to 9 are chosen. One digit is placed in each box to complete the two 2-digit numbers shown. The result of subtracting the bottom number from the top number is calculated. How many of the possible results are positive?

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-014.jpg?height=144&width=179&top_left_y=218&top_left_x=1380)
(A) 36
(B) 32
(C) 30
(D) 34
(E) 38
We begin by calling the missing digits $A$ and $B$, as shown. Each of the digits $A$ and $B$ is chosen from the digits 1 to 9 and $A \neq B$.

The top 2-digit number $5 A$ is at most 59 .

| $5 \boxed{A}$ |
| ---: |
| $-B 5$ |

Thus, $B$ cannot equal $6,7,8$, or 9 since the result of subtracting the bottom number from the top would be negative.

If $B=5$, then the bottom number is 55 and for the result to be positive, $A$ could equal 6,7 , 8 , or 9 (with the results being $1,2,3$, and 4 respectively).

Thus, there are 4 possible positive results in this case.

If $B=4$, then the bottom number is 45 and for the result to be positive, $A$ could equal each of the digits from 1 to 9 with the exception of 4 (since $A \neq B$ ).

In this case, the results of subtracting the bottom number from the top are $6,7,8,10,11,12$, 13 , and 14 respectively. Thus, there are 8 possible positive results when $B=4$.

If $B=3$, then the bottom number is 35 and for the result to be positive, $A$ could equal 1,2 , $4,5,6,7,8$, or 9 (with the results being $16,17,19,20,21,22,23$, and 24 respectively).

Thus, there are 8 possible positive results when $B=3$.

Similarly, when $B=2$ there are 8 possible positive results and when $B=1$ there are 8 possible positive results (and each of these results is different from any of the other results).

In total, the number of possible results that are positive is $4+(8 \times 4)=36$.

ANSWER: (A)