Let S be the set \1, 2, 3, …, 19\. For a, b \in S, define a \succ b to mean that either 0 \lt a - b ≤ 9 or 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?

2013 AMC 12A Problem 20 — Solution

2013 AMC 12A Problem 20

Below is the professionally curated solution for Problem 20 of the 2013 AMC 12A, 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.

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Concepts:modular arithmetic basic counting

Difficulty rating: 2220

20.

Let $S$ be the set $\{1, 2, 3, \ldots, 19\}.$ For $a, b \in S,$ define $a \succ b$ to mean that either $0 \lt a - b \le 9$ or $b - a \gt 9.$ How many ordered triples $(x, y, z)$ of elements of $S$ have the property that $x \succ y,$ $y \succ z,$ and $z \succ x?$

$810$

$855$

$900$

$950$

$988$

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.

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