Let S be the set {1, 2, 3, …, 19}. For a, b ∈ S, define a b to mean that either 0 9. How many ordered triples (x, y, z) of elements of S have the property that x y, y z, and z x?

2013 AMC 12A Problem 20

Attempt Problem 20 of the 2013 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

20.

Let

S

be the set

\{1, 2, 3, \ldots, 19\}.

For

a,

b \in S,

define

a \succ b

to mean that either

0 \lt a - b \le 9

b - a \gt 9.

How many ordered triples

(x, y, z)

of elements of

S

have the property that

x \succ y,

y \succ z,

and

z \succ x?

810

855

900

950

988

Answer: B

Concepts:modular arithmetic basic counting

Difficulty rating: 2220

Small hint:

Work modulo

19;

then

a \succ b

means

0 \lt (a - b) \bmod 19 \le 9

Big hint:

Fix

x

(

19

ways); let

y = x + i

with

1 \le i \le 9,

then

z

has

i

options

Solution:

Reading the elements modulo

19,

the relation

a \succ b

holds exactly when

0 \lt (a - b) \bmod 19 \le 9.

There are

19

choices for

x.

Once

x

is fixed, take

y = x + i

for some

1 \le i \le 9.

Then

z

must satisfy

x + 10 \le z \le x + 9 + i,

giving

i

choices.

The total is

19(1 + 2 + \cdots + 9) = 19\cdot 45

= 855.

Thus, the correct answer is B.