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2014 AMC 10B Problem 24

2014 AMC 10B Problem 24AMC 10 Step by Step

Name: Date: September 13, 2026

Arrangements with Restrictions · difficulty 4 of 5 · about 8 min · position P21-25

The numbers 1, 2, 3, 4, 5 are to be arranged in a circle. An arrangement is bad if it is not true that for every $n$ from $1$ to $15$ one can find a subset of the numbers that appear consecutively on the circle that sum to $n$ . Arrangements that differ only by a rotation or a reflection are considered the same. How many different bad arrangements are there?

A)
$1$
B)
$2$
C)
$3$
D)
$4$
E)
$5$

Answer

Key insight

A consecutive block and its complement are both blocks summing to n and 15 - n, so only sums 6 and 7 can fail; check adjacency conditions.

Solution

The numbers not in a consecutive block are themselves a consecutive block, and the two sums add to $15$ . So if $n$ is achievable, so is $15 - n$ . The sums $1$ through $5$ are single numbers and $15$ is the whole circle, so an arrangement is good exactly when both $6$ and $7$ are achievable.

Up to rotation and reflection there are $\frac{4!}{2} = 12$ arrangements. Fix $1$ at the top.

Ways to make $6$ : adjacent pair $1,5$ ; adjacent pair $2,4$ ; or the block $\{1,2,3\}$ .
Ways to make $7$ : adjacent pair $2,5$ ; adjacent pair $3,4$ ; or the block $\{1,2,4\}$ .

Arrangements missing $6$ . Then $1$ 's two neighbors come from $\{2,3,4\}$ . If they are $2,3$ , the block $\{1,2,3\}$ appears. If they are $3,4$ , the far pair is $\{2,5\}$ and the circle is $1,3,2,5,4$ (block $1,3,2$ sums to $6$ ) or $1,3,5,2,4$ ( $2,4$ adjacent); both make $6$ . If they are $2,4$ , the far pair is $\{3,5\}$ : $1,2,3,5,4$ contains the block $1,2,3$ , but $1,2,5,3,4$ has no $1{-}5$ adjacency, no $2{-}4$ adjacency, and no $\{1,2,3\}$ block. One bad arrangement.

Arrangements missing $7$ . Then $5$ 's neighbors come from $\{1,3,4\}$ and $3,4$ are not adjacent. Checking the possibilities the same way, only $1,3,2,4,5$ survives: $2$ touches $3,4$ ; $3$ touches $1,2$ ; and $1,2,4$ are not consecutive. One bad arrangement.

Total: $1 + 1 = 2$ . The answer is $\boxed{\textbf{(B)}\ 2}$ .

Why this works

On a circle, complements of arcs are arcs; that symmetry halves the work, leaving only two target sums. With only twelve arrangements, careful adjacency reasoning (or a quick full listing) finishes. The lesson: before enumerating, find the structural fact that shrinks what must be checked.

Alternative approach

List all twelve arrangements with $1$ fixed, $2$ before $5$ clockwise (to kill reflections), and for each record whether $6$ and $7$ are block sums. Only $1,2,5,3,4$ (missing $6$ and $9$ ) and $1,3,2,4,5$ (missing $7$ and $8$ ) fail.

The trap

Testing all fifteen sums for each of the twelve arrangements, or forgetting to quotient by reflection and reporting a doubled count.

Common mistakes

Testing all fifteen sums for each of the twelve arrangements, or forgetting to quotient by reflection and reporting a doubled count.
Missing the three-element blocks ( $1,2,3$ or $1,2,4$ ) as ways to make $6$ or $7$ , which wrongly marks extra arrangements as bad.

Techniques

Organized listing / direct enumeration · Exploit symmetry to reduce work or pair up objects