2013 AMC 10B Problem 22

Question

The regular octagon

ABCDEFGH

has its center at

J.

Each of the vertices and the center are to be associated with one of the digits

1

through

9,

with each digit used once, in such a way that the sums of the numbers on the lines

AJE,

BJF,

CJG,

and

DJH

are all equal. In how many ways can this be done?

384

576

1152

1680

3456

The center digit contributes to all four line sums

After choosing the center, pair the remaining digits into equal-sum opposite pairs

Let

S

be defined as:

\begin{aligned} S &= A+J+E \\ &=B+J+F \\ &=C+J+G \\ &=D+J+H \end{aligned}

\begin{aligned} 4S &= A+B+C+D+E \\ &\quad+F+G+H+4J \end{aligned}

4S = 45+3J

45+3J \equiv 0 \mod 4

3J \equiv 3 \mod 4

J \equiv 1 \mod 4

This means that

J=1,5,9.

From here, let’s assume

J=1.

We will see that the other cases are similar enough to omit.

If

J=1,

then we know that the pairs of numbers that satisfy the equality above are:

2+9 = 3+8 = 4+7 = 5+6

There are

4!

ways to distribute the pairs over the four groups, and then

2^4

ways for these groups to swap elements (i.e.

2+9\iff 9+2

).

Now, if we look at the

J=5

and

J=9

cases, we see a similar pattern in the number of groupings and swaps. As such, we have:

3\cdot 4! \cdot 2^4 = 1152

possibilities.

Thus, the correct answer is C .

Problem 22 in other years