Problem 958

Suppose that the length, or the width, or the height of the
rectangular prism is equal to 5.

The product of 5 with any of the remaining digits has a units (ones)
digit that is equal to 5 or it is equal to 0.

This means that if the length, or the width, or the height of the
rectangular prism is equal to 5, then at least one of the two-digit
integers (the area of a face) has a units digit that is equal to 5 or
0.

However, 0 is not a digit that can be used, and each digit from 1 to 9
is used exactly once (that is, 5 cannot be used twice), and so it is not
possible for one of the dimensions of the rectangular prism to equal
5.

Thus, the digit 5 occurs in one of the two-digit integers (the area of a
face).

The digit 5 cannot be the units digit of the area of a face, since this
would require that one of the dimensions be 5.

Therefore, one of the areas of a face has a tens digit that is equal to
5.

The two-digit integers with tens digit 5 that are equal to the product
of two different one-digit integers (not equal to 5) are $54=6\times 9$ and $56=7\times 8$.

Suppose that two of the dimensions of the prism are 7 and 8, and so one
of the areas is 56.

In this case, the digits $5,6,7$, and
8 have been used, and so the digits $1,2,3,4$, and 9 remain.

Which of these digits is equal to the remaining dimension of the
prism?

It cannot be 1 since the product of 1 and 7 does not give a two-digit
area, nor does the product of 1 and 8.

It cannot be 2 since the product of 2 and 8 is 16 and the digit 6 has
already been used.

It cannot be 3 since $3\times 7=21$
and $3\times 8=24$, and so the areas
of two faces share the digit 2.

It cannot be 4 since $4\times 7=28$
and the digit 8 has already been used.

Finally, it cannot be 9 since $9\times7=63$ and the digit 6 has already
been used.

Therefore, it is not possible for 7 and 8 to be the dimensions of the
prism, and thus 6 and 9 must be two of the three dimensions.

Using a similar systematic check of the remaining digits, we determine
that 3 is the third dimension of the prism.

That is, when the dimensions of the prism are $3,6$ and 9, the areas of the faces are
$3\times 6=18$, $3\times 9=27$, and $6\times9=54$, and we may confirm that each
of the digits from 1 to 9 has been used exactly once.

Since the areas of the faces are 18, 27 and 54, the surface area of the
rectangular prism is $2\times(18+27+54)$ or $2\times99=198$.