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

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== Solution 1 ==
== Solution 1 ==
let 0<=x<y<z<=8;
Let <imath>0 \le x<y<z \le 8</imath>;
x cannot be 0 because it makes xy>z --> 0>z;  
<imath>x</imath> cannot be <imath>0</imath> because it makes <imath>xy>z</imath> <imath>\rightarrow</imath> <imath>0>z</imath>;  
x cannot be 1 because it makes xy>z ---> y>z;
<imath>x</imath> cannot be <imath>1</imath> because it makes <imath>xy>z</imath> <imath>\rightarrow</imath> <imath>y>z</imath>;


x=2, y=3, z can be 4, 5 but not others;  
<imath>x=2</imath>, <imath>y=3</imath>, <imath>z</imath> can be <imath>4</imath>, <imath>5</imath> but not others;  
x=2, y=4, z can be 5, 6, 7;
<imath>x=2</imath>, <imath>y=4</imath>, <imath>z</imath> can be <imath>5</imath>, <imath>6</imath>, <imath>7</imath>;
x=2, y=5, z can be 6, 7, 8;
<imath>x=2</imath>, <imath>y=5</imath>, <imath>z</imath> can be <imath>6</imath>, <imath>7</imath>, <imath>8</imath>;
x=2, y=6, z can be 7, 8;
<imath>x=2</imath>, <imath>y=6</imath>, <imath>z</imath> can be <imath>7</imath>, <imath>8</imath>;
x=2, y=7, z can be 8;
<imath>x=2</imath>, <imath>y=7</imath>, <imath>z</imath> can be <imath>8</imath>;
for x=2, total 11 cases;  
for <imath>x=2</imath>, total <imath>11</imath> 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;
Similarly, for <imath>x=3</imath>, <imath>y=4</imath>, <imath>5</imath>, <imath>6</imath>, <imath>7</imath>, total <imath>10</imath> cases; for <imath>x=4</imath>, <imath>y=5</imath>, <imath>6</imath>, <imath>7</imath>, total <imath>6</imath> cases; <imath>x=5</imath>, <imath>y=6</imath>, <imath>7</imath>, <imath>3</imath> cases; <imath>x=6</imath>, <imath>y=7</imath>, <imath>z=8</imath>, <imath>1</imath> cases;


total = 11 + 10 +6 +3 +1 = 31. permutate x, y, z for ordered triple, it is 31*6=186, C.
Total <imath>= 11 + 10 + 6 + 3 + 1 = 31</imath>. Permutate <imath>x</imath>, <imath>y</imath>, <imath>z</imath> for ordered triple, it is <imath>31 \cdot 6=186</imath>, <imath>\fbox{C}</imath>.




~imagination
~imagination
~mathcantcount1plus1is3 (latex stuff)
Note: x also cannot equal 0 because the problem specifies that x, y, and z are all positive integers.


==Solution 2==
==Solution 2==
Line 48: Line 55:
Note that if <imath>x,y,z>2,</imath> then this must hold, as if it didn't, we would need, WLOG, <imath>x>3y,</imath> which would imply <imath>x>8.</imath> Therefore, there are at least <imath>6\times 5\times 4=120</imath> solutions. However, we must have <imath>x,y,z\ge2,</imath> as if any variable is <imath>0,</imath> this clearly doesn't work, and <imath>x=1</imath> gives <imath>y>z</imath> and <imath>z>y,</imath> which is impossible. Therefore, there are at most <imath>7\times6\times5=210</imath> solutions. The only choice that works from here is <imath>\boxed{\textbf{(C)}~186}</imath>
Note that if <imath>x,y,z>2,</imath> then this must hold, as if it didn't, we would need, WLOG, <imath>x>3y,</imath> which would imply <imath>x>8.</imath> Therefore, there are at least <imath>6\times 5\times 4=120</imath> solutions. However, we must have <imath>x,y,z\ge2,</imath> as if any variable is <imath>0,</imath> this clearly doesn't work, and <imath>x=1</imath> gives <imath>y>z</imath> and <imath>z>y,</imath> which is impossible. Therefore, there are at most <imath>7\times6\times5=210</imath> solutions. The only choice that works from here is <imath>\boxed{\textbf{(C)}~186}</imath>
~krithikrokcs
~krithikrokcs
==Solution 5(Compl. Counting)==
The total number of cases is <imath>7*6*5=210</imath>. Note that no number can be 1, since that would result in cases like <imath>z>y</imath> and <imath>z<y</imath> which is impossible.
We can try complementary counting. Since in this problem x,y,z cases are symmetrical, we first consider cases where <imath>xy<z</imath>. The cases are:
<imath>(x,y,z): (2,3,6) (2,3,7) (2,3,8) (2,4,8)</imath>. By symmetry, the cases that don't work amounts to <imath>4*6=24</imath> cases. Therefore the total number of cases that work is <imath>210-24=186</imath>, which is <imath>\boxed{\textbf{(C)}~186}</imath>.
~backtosq-1
== Solution 6 (Symmetry + Complementary Counting) ==
Since <imath>\{ x,y,z \} \subset \{ 1,2,3,4,5,6,7,8 \}</imath>, WLOG let <imath>1\le x<y<z\le 8</imath> (<imath>\times 3! =6</imath>). Note that the lower bounds make the conditions <imath>y<xz\le 1z</imath> and <imath>x<zy<z</imath> trivial. In addition, <imath>x=1</imath> always fails, since it would force <imath>z<xy=y</imath>, a contradiction. Thus, it remains exactly to check <imath>2\le x<y<z\le 8<xy</imath>.
The only failure cases with <imath>xy\le 8</imath> can occur when <imath>(x,y)=(2,3),(2,4)</imath>, for which we have corresponding <imath>z\in \{ 6,7,8 \},\{ 8 \}</imath>. The total number of cases is <imath>\binom{7}{3}=35</imath>. Thus, the number of good cases is <imath>35-4=31</imath>, and after multiplying by order we get <imath>3!\cdot 31=\boxed{\textbf{(C)}~186}</imath>.
~imosilver


==Video Solution 1 by OmegaLearn==
==Video Solution 1 by OmegaLearn==
https://youtu.be/OPQbAyYZBtA
https://youtu.be/OPQbAyYZBtA
==Video Solution by SpreadTheMathLove==
https://www.youtube.com/watch?v=dAeyV60Hu5c


==See Also==
==See Also==

Latest revision as of 22:37, 11 November 2025

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)



Note: x also cannot equal 0 because the problem specifies that x, y, and z are all positive integers.

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

Solution 5(Compl. Counting)

The total number of cases is $7*6*5=210$. Note that no number can be 1, since that would result in cases like $z>y$ and $z<y$ which is impossible.

We can try complementary counting. Since in this problem x,y,z cases are symmetrical, we first consider cases where $xy<z$. The cases are: $(x,y,z): (2,3,6) (2,3,7) (2,3,8) (2,4,8)$. By symmetry, the cases that don't work amounts to $4*6=24$ cases. Therefore the total number of cases that work is $210-24=186$, which is $\boxed{\textbf{(C)}~186}$. ~backtosq-1

Solution 6 (Symmetry + Complementary Counting)

Since $\{ x,y,z \} \subset \{ 1,2,3,4,5,6,7,8 \}$, WLOG let $1\le x<y<z\le 8$ ($\times 3! =6$). Note that the lower bounds make the conditions $y<xz\le 1z$ and $x<zy<z$ trivial. In addition, $x=1$ always fails, since it would force $z<xy=y$, a contradiction. Thus, it remains exactly to check $2\le x<y<z\le 8<xy$.

The only failure cases with $xy\le 8$ can occur when $(x,y)=(2,3),(2,4)$, for which we have corresponding $z\in \{ 6,7,8 \},\{ 8 \}$. The total number of cases is $\binom{7}{3}=35$. Thus, the number of good cases is $35-4=31$, and after multiplying by order we get $3!\cdot 31=\boxed{\textbf{(C)}~186}$.

~imosilver

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)
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Problem 17 		Followed by
Problem 19
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All AMC 12 Problems and Solutions

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