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Difference between revisions of "2024 AMC 12B Problems/Problem 10"

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<math>\textbf{Case 4:}</math>
 
<math>\textbf{Case 4:}</math>
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Since <math>x>1, z>7</math>, the max is <math>z</math> and the min is <math>1</math>. Thus <math>z-1 = 7 \implies z = 8</math>. Now the median is between <math>5.2</math> and <math>6.2</math>, so <math>x</math> or <math>y</math> must take the value <math>6. Our sum is now </math>24.8+6+8 = 38.8<math>, and adding the last number has to equal </math>45<math>. Thus the last number is </math>6.2<math>, and this yields one solution: </math>(6,6.2,8)<math>.
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Therefore we've found three solutions. These cases encompass all possible real values of </math>x<math> and </math>z<math>, so we know we've left nothing out. The answer is </math>\box{C}$
  
 
~KingRavi
 
~KingRavi

Revision as of 02:26, 14 November 2024

Problem 10

A list of 9 real numbers consists of $1$, $2.2$, $3.2$, $5.2$, $6.2$, $7$, as well as $x, y,z$ with $x\leq y\leq z$. The range of the list is $7$, and the mean and median are both positive integers. How many ordered triples $(x,y,z)$ are possible?

$\textbf{(A) }1 \qquad\textbf{(B) }2 \qquad\textbf{(C) }3 \qquad\textbf{(D) }4 \qquad\textbf{(E) \text{infinitely many}}\qquad$

Solution

It is easiest to do casework on the range. We have four possible cases:

$x<1$, $z<7$

$x>1$, $z<7$

$x<1$, $z>7$

$x>1$, $z>7$

These encapsulate all possible values of $x$ and $z$ we can choose, so we're not leaving anything out. As we'll see, the problem will become simpler this way. Since we want integer mean, first note the sum of the values given to be $24.8$: when we add $x,y,z$, it must be a multiple of $9$ to yield an integer mean. Also remember that the median of an increasing list of $9$ numbers is the fifth number.

$\textbf{Case 1:}$

Since $x<1$, $z<7$, $x$ must be the minimum of this list of numbers and $7$ is the max: so the range is $7-x$. This must be equal to $7$, so $x=0$. Now we try to make the median an integer. With $x=0$, the fifth smallest number currently is $5.2$, which is clearly not an integer. But if we stick $y$ or $z$ in between $3.2$ and $5.2$, then the median will be an integer. So let one of $y,z$ be $4$: then the sum of all the numbers we have so far is $28.8$. To make this a multiple of $9$, the last number has to be $7.2$ to add up to $36$. However, we said that $z<7$, so this violates the bounds of this case. So then we set $y=5$, so our sum is $29.9$: then $z=36-29.9 = 6.2$. Here everything checks out, so for this case there is one solution: $(0,5,6.2)$.

$\textbf{Case 2:}$

Since $x>1$ and $z<7$, the minimum is $1$ and the maximum is $7$. Thus the range is always $6$, but we need the range to be $7$, so this case has no solutions.

$\textbf{Case 3:}$

We have $x<1, z>7$. Our minimum is $x$ and our maximum is $z$, so the range is $z-x = 7$. Since $x<1$, the median still falls on $5.2$, so $y$ can be either $4$ or $5$ to make the median an integer. For now, suppose $y=4$. Our sum is now $28.8+x+z$, which we set to the nearest multiple of $9:$ $36$. Thus $x+z = 7.2$. Now we have a system of equations: we add them to get $2z = 14.2 \implies z = 7.1 \implies x = 0.1$. Since the values of $z$ and $x$ satisfy our case, this is a possible solution.

Now suppose $y=5$. Our sum is $29.8+x+z$, which must be equal to $36$, so $x+z = 6.2$. Adding the equations, we have $2z = 13.2 \implies z = 6.6$. But this violates our assumption that $z>7$, so this has no solutions.

Therefore this case has one solution: $(0.1,4,7.1)$.

$\textbf{Case 4:}$

Since $x>1, z>7$, the max is $z$ and the min is $1$. Thus $z-1 = 7 \implies z = 8$. Now the median is between $5.2$ and $6.2$, so $x$ or $y$ must take the value $6. Our sum is now$24.8+6+8 = 38.8$, and adding the last number has to equal$45$. Thus the last number is$6.2$, and this yields one solution:$(6,6.2,8)$.



Therefore we've found three solutions. These cases encompass all possible real values of$ (Error compiling LaTeX. Unknown error_msg)x$and$z$, so we know we've left nothing out. The answer is$\box{C}$

~KingRavi

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