2025 AMC 12A Problems/Problem 18

Problem 19

How many ordered triples $(x, y, z)$ of different positive integers less than or equal to $8$ satisfy $xy > z$ , $xz > y$ , and $yz > x$ ?

$\textbf{(A)}~36 \qquad \textbf{(B)}~84 \qquad \textbf{(C)}~186 \qquad \textbf{(D)}~336 \qquad \textbf{(E)}~486$

Solution 1

let 0<=x<y<z<=8; x cannot be 0 because it makes xy>z --> 0>z; x cannot be 1 because it makes xy>z ---> y>z;

x=2, y=3, z can be 4, 5 but not others; x=2, y=4, z can be 5, 6, 7; x=2, y=5, z can be 6, 7, 8; x=2, y=6, z can be 7, 8; x=2, y=7, z can be 8; for x=2, total 11 cases;

similarly, for x=3,y=4, 5, 6, 7, total 10 cases; for x=4, y =5, 6, 7, total 6 cases; x=5, y=6, 7, 3 cases; x=6, y=7, z=8, 1 cases;

total = 11 + 10 +6 +3 +1 = 31. permutate x, y, z for ordered triple, it is 31*6=186, C.

Solution 2

We may use complementary counting:

For now, assume $x\leq y\leq z$ .

First note that no number can be 0, as it would imply $0y>z$ . Similarly, no number can be 1, as it would imply $1y>z$ . So we only need to consider numbers between 2 and 8, inclusive.

Consider when $xy\leq z$ . This implies $xy\leq 8$ . Some quick calculations gives us the products $2\cdot2,2\cdot3,2\cdot4$ . We may now calculate the number of times each happens (we are no longer assuming $x\leq y\leq z$ ):