2025 AMC 12A Problems/Problem 18: Difference between revisions

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Problem 18

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 $0 \le x<y<z \le 8$ ; $x$ cannot be $0$ because it makes $xy>z$ $\rightarrow$ $0>z$ ; $x$ cannot be $1$ because it makes $xy>z$ $\rightarrow$ $y>z$ ;

$x=2$ , $y=3$ , $z$ 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 $31 \cdot 6=186$ , $\fbox{C}$ .

~imagination

~mathcantcount1plus1is3 (latex stuff)

Solution 2

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.

We may use complementary counting:

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 $x\leq y\leq z$ ):

$2\cdot2$ : This case is invalid as it asks for distinct integers.
$2\cdot3$ : $2,3,(n\geq6)$ . Then we have $6\cdot3=18$ cases.
$2\cdot4$ : $2,4,(n\geq8)$ . Then we have $6\cdot1=6$ cases.

In total, there are $7\cdot6\cdot5=210$ total cases, so our final answer is $210 - (18 + 6) = \boxed{186}$ .

~SilverRush

Solution 3

Note that if $x,y,z \ge 3$ , all pairs work - hence, we have $\binom{6}{3} \cdot 6 = 120$ pairs. Now, note that if $x=1$ , we get $y>z$ and $z>y$ , contradiction. Therefore, assume $x=2$ (since we said $x < 3, x \neq 1$ ). Note that we need $2y>z>\frac{y}{2}$ . WLOG assume $y>z$ , we get there are $11$ pairs that work (we just need $2z>y$ ): $(y,z) = (8,5), (8,6), (8,7), (7,4), (7,5), (7,6), (6,4), (6,5), (5,3), (5,4), (4,3)$ . With $x=2$ , we can re-arrange these in $3!=6$ ways each, hence the answer is just $120 + 6 \cdot 11 = \boxed{186}$ .

~ScoutViolet

Solution 4 (Bounding)

Note that if $x,y,z>2,$ then this must hold, as if it didn't, we would need, WLOG, $x>3y,$ which would imply $x>8.$ Therefore, there are at least $6\times 5\times 4=120$ solutions. However, we must have $x,y,z\ge2,$ as if any variable is $0,$ this clearly doesn't work, and $x=1$ gives $y>z$ and $z>y,$ which is impossible. Therefore, there are at most $7\times6\times5=210$ solutions. The only choice that works from here is $\boxed{\textbf{(C)}~186}$ ~krithikrokcs

Video Solution 1 by OmegaLearn

https://youtu.be/OPQbAyYZBtA

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=dAeyV60Hu5c

@@ Line 53: / Line 53: @@
 ==Video Solution 1 by OmegaLearn==
 https://youtu.be/OPQbAyYZBtA
+==Video Solution by SpreadTheMathLove==
+https://www.youtube.com/watch?v=dAeyV60Hu5c
 ==See Also==

2025 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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