The digits from 1 to 9 are each used exactly once to write three one-digit integers and three two-digit integers. The one-digit integers are equal to the length, width and height of a rectangular prism. The two-digit integers are equal to the areas of the faces of the same prism. What is the surface area of the rectangular prism?
(A) 176
(B) 184
(C) 186
(D) 198
(E) 212

## Grade 7
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 $9 \times 7=63$ 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 $6 \times 9=54$, 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 $2 \times 99=198$.

ANSWER: (D)

#