2014 AMC 12B Problem 18

Question

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?

1

2

3

4

5

Answer: B

Concepts:circular arrangements complementary counting casework

Difficulty rating: 2150

Small hint:

Sums

1

through

5

are always achievable with a single number

Big hint:

If a consecutive block sums to

n,

its complement sums to

15-n,

so only

6

and

7

can fail

Solution:

Any single number covers sums

1

through

5.

If a consecutive block sums to

n,

the remaining numbers form a consecutive block summing to

15 - n,

so sums

10

through

14

are automatically covered as well. Thus an arrangement is bad only if it fails to produce

6

or

7.

If

6

cannot be formed, rotate and reflect so the order is

1bc5e.

The pair

\{b,c\}

cannot be

\{2,3\}

or

\{2,4\},

so

e=2;

avoiding the block

213

then forces

14352.

If

7

cannot be formed, write the order as

2bc5e.

Now

\{b,c\}

cannot be

\{3,4\}

or

\{1,4\},

so

e=4,

and avoiding

421

forces

23154.

These are the only two bad arrangements up to rotation and reflection.

Thus, the correct answer is B.

2014 AMC 12B Problem 18

Answer: B

Difficulty rating: 2150

Small hint:

Big hint:

Solution:

Problem 18 in other years