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

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total = 11 + 10 +6 +3 +1 = 31. permutate x, y, z for ordered triple, it is 31*6=186, C.
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 <imath>x\leq y\leq z</imath>.
First note that no number can be 0, as it would imply <imath>0y>z</imath>. Similarly, no number can be 1, as it would imply <imath>1y>z</imath>. So we only need to consider numbers between 2 and 8, inclusive.
Consider when xy\leq z. This implies <imath>xy\leq 8</imath>. Some quick calculations gives us the products <imath>2\cdot2,2\cdot3,2\cdot4</imath>. We may now calculate the number of times each happens (we are no longer assuming <imath>x\leq y\leq z</imath>):
\begin{itemize}
\item <imath>2\cdot2</imath>: This case is invalid as it asks for distinct integers.
\item <imath>2\cdot3</imath>: <imath>2,3,(n\geq6)</imath>. Then we have <imath>6\cdot3=18</imath> cases.
\item <imath>2\cdot4</imath>: <imath>2,4,(n\geq8)</imath>. Then we have <imath>6\cdot1=6</imath> cases.
\end{itemize}
In total, there are <imath>7\cdot6\cdot5=210</imath> total cases, so our final answer is <imath>210 - (18 + 6) = \boxed{186}</imath>.
~SilverRush

Revision as of 17:23, 6 November 2025

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 $x\leq y\leq z$): \begin{itemize} \item $2\cdot2$: This case is invalid as it asks for distinct integers. \item $2\cdot3$: $2,3,(n\geq6)$. Then we have $6\cdot3=18$ cases. \item $2\cdot4$: $2,4,(n\geq8)$. Then we have $6\cdot1=6$ cases. \end{itemize} In total, there are $7\cdot6\cdot5=210$ total cases, so our final answer is $210 - (18 + 6) = \boxed{186}$. ~SilverRush

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