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2013 AMC 10B Problem 22

2013 AMC 10B Problem 22AMC 10 Step by Step

Name: Date: September 13, 2026

Arrangements with Restrictions · difficulty 4 of 5 · about 7 min · position P21-25

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?

A)
$384$
B)
$576$
C)
$1152$
D)
$1680$
E)
$3456$

Answer

Key insight

The center is on every line, so opposite pairs need equal sums; 45-J divisible by 4 forces J in {1,5,9}, then 4!*2^4 each.

Solution

Each of the four lines passes through the center $J$ , so its sum is $J$ plus the sum of a pair of opposite vertices. The lines have equal sums exactly when the four opposite pairs $\{A, E\}, \{B, F\}, \{C, G\}, \{D, H\}$ all have the same pair-sum $s$ .

The eight vertex numbers are everything except $J$ , so $4s = 45 - J$ . Thus $45 - J$ must be a multiple of $4$ , which happens for $J = 1, 5, 9$ only.

$J = 1$ : $s = 11$ , pairs $\{2,9\}, \{3,8\}, \{4,7\}, \{5,6\}$ .
$J = 5$ : $s = 10$ , pairs $\{1,9\}, \{2,8\}, \{3,7\}, \{4,6\}$ .
$J = 9$ : $s = 9$ , pairs $\{1,8\}, \{2,7\}, \{3,6\}, \{4,5\}$ .

In every case the eight remaining digits split into four pairs in exactly one way. Assign the four pairs to the four lines: $4! = 24$ ways. Within each line, decide which end gets which number: $2^4 = 16$ ways. Each case contributes $24 \cdot 16 = 384$ labelings.

Total: $3 \cdot 384 = 1152$ .

The answer is $\boxed{\textbf{(C)}\ 1152}$ .

Why this works

A shared element (the center) drops out of every comparison, reducing "four equal line sums" to "four equal pair sums," a partition problem controlled by a divisibility condition. Once the pairs are known, the count is pure arrangement: which pair goes where, and which way round.

The trap

Forgetting that each pair can be flipped (factor 2^4), or that the four pairs can be assigned to the four lines in 4! ways.

Common mistakes

Forgetting that each pair can be flipped (factor 2^4), or that the four pairs can be assigned to the four lines in 4! ways.
Counting only one value of $J$ (getting $384$ , choice (A)) or including $J$ values like $3$ or $7$ that leave $45 - J$ not divisible by $4$ .

Techniques

Split into exhaustive cases and handle each · Set up the equation/formula and compute; no special trick needed