The digits 1, 2, 3, 4, 5, and 6 are each placed in one of the boxes so that the resulting product is correct. If each of the six digits is used exactly once, the digit represented by "?" is
(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

![](https://cdn.mathpix.com/cropped/2024_04_20_46ead6524a8d61e21c51g-086.jpg?height=219&width=247&top_left_y=2097&top_left_x=1427)
We will present a complete consideration of all of the cases. The answer can be obtained more easily in a trial and error fashion.

First, we rewrite the equation putting letters in each of the boxes

| $A \quad B$ |
| ---: |
| $\times \quad C$ |
| $D E \quad F$ |

We want to replace $A, B, C, D, E$, and $F$ by the digits 1 through 6 .

## Could $C$ be 1 ?

If $C$ was 1 , then $B$ and $F$ would have to be same digit, which is impossible since all of the digits are different. Therefore, $C$ cannot be 1 .

Could $C$ be 5 ?

If $C$ was 5 , then if $B$ were odd, $F$ would also be 5 , which would be impossible. If $C$ was 5 and $B$ even, then $F$ would have to be 0 , which is also impossible. Therefore, $C$ cannot be 5 .

Could $C$ be 6 ?

If $C$ was 6 , let us list the possibilities for $B$ and the resulting value of $F$ :

| $B$ | $F$ |
| :--- | :--- |
| 1 | 6 |
| 2 | 2 |
| 3 | 8 |
| 4 | 4 |
| 5 | 0 |

$$
\begin{aligned}
& \text { Impossible - two 6's } \\
& \text { Impossible - two 2's } \\
& \text { Impossible - no } 8 \\
& \text { Impossible - two 4's } \\
& \text { Impossible - no } 0
\end{aligned}
$$

Therefore, $C$ cannot be 6 .

Could $C$ be 4 ?

If $C$ was 4 , using a similar chart, we can see that $B$ must be 3 and $F$ must be 2 . We will consider this possibility later.

## Could $C$ be 3?

If $C$ was 4 , using a similar chart, we can see that $B$ must be 2 or 4 and $F$ must be thus 6 or 2 . Could $C$ be 2?

If $C$ was 2, using a similar chart, we can see that $B$ must be 3 and $F$ must be 6 .

Let us consider the case of $C=2$. In this case, we have

| $A 3$ |
| ---: |
| $\times \quad 2$ |
| $D \quad 6$ |

Since the product has three digits, and the possibilities remaining for $A$ are 1,4 or 5 , then $A$ must be 5. However, this gives a product of $53 \times 2=106$, which is impossible.

Similarly, trying the case of $C=4$, we are left with the possibilities for $A$ being 1,5 , or 6 , none of which work.

Therefore, $C$ must be 3, since this is the only possibility left. We should probably check that we can actually get the multiplication to work, though!

Trying the possibilities as above, we can eventually see that $54 \times 3=162$ works, and so $C=3$.

ANSWER: (B)