2024 AMC 10B Problems/Problem 15

The following problem is from both the 2024 AMC 10B #15 and 2024 AMC 12B #10, so both problems redirect to this page.

Problem

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

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

Solution 1 (AMC 10)

The sum of the six existing numbers is $24.8$ . For the mean to be an integer, the sum of the remaining three numbers $x,y,z$ and the remaining six must be divisible by $9$ ; thus $x+y+z$ must equal either $2.2,11.2,20.2$ . However, one of $x,y,z$ must be the median since the middle four numbers in the original set are all non-integral. As a result, we can eliminate $2.2$ since the sum would be too small to allow for one of $x,y,z$ to be the median (notice that negative numbers are not permitted since the range would be greater than $7$ ). This leaves two cases.

For the $x+y+z=20.2$ case, notice that all three must be less than or equal to $8$ due to the range restriction. However, also notice that $y$ and $z$ must be on the higher end of the range, meaning that the new median would have to fall between the new middle numbers $5.2$ and $6.2$ . Thus $x=6$ , and we can manipulate the numbers to make the range $7$ by making $z=8$ and thus $y=6.2$ , providing one case in $(6,6.2,8)$ .

For the $x+y+z=11.2$ case, we note that the average of the three would fall near the median, signaling that one of the numbers would be the median, one would fall lesser than the median, and one would reach greater than the median. Thus we let the median be $y$ . Since adding a number to each side of the set would not change the median, we know that the median must fall between the two middle numbers $3.2,5.2$ . Then we find two triples $(0.1,4,7.1)$ and $(0,5,6.2)$ , resulting in $\boxed{\textbf{(C) }3}$ .

~eevee9406

Solution 2

We can start doing casework on the numbers. Notice that the median can either be four, five or six because there has to be one or two numbers on one side. We can start bashing everything out, starting from letting $x=a, y=4, z=7+a$ . Now plugging into the sequence, we get that $a, 1, 2.2, 3.2, 4, 5.2, 6.2, 7, 7+a$ . Notice that the sum of the numbers (not including $x, y, z$ is $24.8$ . To make this mean an integer, $x+y+z$ must have a decimal $0.2$ when added up. Now, plugging in our values into the mean gets us $35.8+2a=9k$ for some integer $k$ . Notice that $0\leq a\leq 1$ , so the only value for $a$ is $0.1$ , so we found our first ordered triplet, $(0.1, 4, 7.1)$ . Let’s make the median $5$ now. Then we have to put $x$ and one side and $z$ on the other to “balance” the numbers, or make it so that there are 4 numbers on each side of $5$ . Now, we can see that $x$ is $0$ , and solving for $z$ , we get that $z=6.2$ . We get our second triple is $(0, 5, 6.2)$ . Now coming to our third case scenario, we use $6$ as the median. Clearly it means that $x$ has to be the median, so that $y$ and $z$ can be on the same side as $x$ . $1, 2.2, 3.2, 5.2, 6, 6.2, y, 7, z$ We will use this diagram. Notice that $z$ has to be $8$ , and solving for $y$ yields $y=6.2$ , so we found our third and last triplet, $(6, 6.2, 8)$ , meaning that $\boxed{C}$ is the right answer.

~EaZ_Shadow

Solution 3 (AMC 12)

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 $x$ and $z$ , so we know we've left nothing out. The answer is $\boxed{\textbf{(C) }3}$

~KingRavi

Solution 3(Just do it✅）

E is too big and AB are too small, use the method of 掐头去尾 and subsequently do some slickback, you get 3 as final answer, 3 Note: The answer to AoPS problems is never, EVER infinitely many. Also, the top part wasn't mine, it was somebody else. ~ PerseverePlayer

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/YqKmvSR1Ckk?feature=shared

~ Pi Academy

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=24EZaeAThuE

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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All AMC 10 Problems and Solutions

2024 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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