2025 AMC 12A Problems/Problem 18

Problem 19

How many ordered triples of different positive integers less than or equal to satisfy , , and ?

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.

~imagination

Solution 2

For now, assume .

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

We may use complementary counting:

Consider when . This implies . Some quick calculations gives us the products . We may now calculate the number of times each happens (we are no longer assuming ):

• : This case is invalid as it asks for distinct integers.
• : . Then we have cases.
• : . Then we have cases.

In total, there are total cases, so our final answer is .

~SilverRush

Solution 3

Note that if , all pairs work - hence, we have pairs. Now, note that if , we get and , contradiction. Therefore, assume (since we said ). Note that we need . WLOG assume , we get there are pairs that work (we just need ): . With , we can re-arrange these in ways each, hence the answer is just .

~ScoutViolet

Solution 4 (Bounding)

Note that if then this must hold, as if it didn't, we would need, WLOG, which would imply Therefore, there are at least solutions. However, we must have as if any variable is this clearly doesn't work, and gives and which is impossible. Therefore, there are at most solutions. The only choice that works from here is

Video Solution 1 by OmegaLearn

https://youtu.be/OPQbAyYZBtA