AMC 10 · 2024 · #15

Grade 6 arithmetic

Archetype Small-Domain Constrained Search

mean-median-mode-rangesystematic-enumerationinterval-arithmetic caseworksystematic-enumerationidentify-subproblems ↑ Prerequisites: mean-median-mode-rangefraction-decimal-conversion

📏 Long solution 💡 3 insights

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?

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

Pick an answer.

(A)

(B)

(C)

(D)

(E)

$text{infinitely many}$

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Toolkit + CCSS Solution

Understand

Restated: A list of $9$ real numbers contains the six fixed values $1, 2.2, 3.2, 5.2, 6.2, 7$ together with three more reals $x \leq y \leq z$. The range of the list is $7$, and both the mean and the median are positive integers. Count the number of ordered triples $(x, y, z)$ that work.

Givens: Six fixed numbers in the list: $1, 2.2, 3.2, 5.2, 6.2, 7$; their sum is $24.8$; Three free reals $x \leq y \leq z$; Range of the $9$-number list $= 7$; Mean is a positive integer; median (5th number when sorted) is a positive integer; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) infinitely many

Unknowns: How many ordered triples $(x, y, z)$ satisfy all four conditions

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities

There are three independent conditions — range, mean-integer, median-integer — and four ways the min/max of the list can fall on $\{1, 7\}$ versus $\{x, z\}$. Tool #7 (Identify Subproblems) makes the casework explicit by partitioning on "who is min, who is max". Tool #2 (Systematic List) then walks the three live cases (the fourth, $\min = 1, \max = 7$, has range $6$ and is killed instantly). Within each case, the mean-integer condition pins $x + y + z \bmod 1$ to $.2$, the range condition pins one of $x, z$, and the median-integer condition pins the remaining free coordinate. Tool #3 (Eliminate Possibilities) does the final filtering — once we have a candidate, plug back and check all four conditions; tossing the ones that fail.

Execute — Answer: C

#7 Identify Subproblems 6.SP.B.5 Step 1

Translate the conditions into algebra.
Sum of the six fixed numbers: $1 + 2.2 + 3.2 + 5.2 + 6.2 + 7 = 24.8$.
For the mean to be an integer $M$, we need $24.8 + x + y + z = 9M$.
So $x + y + z \in \{2.2, 11.2, 20.2, 29.2, \dots\}$.
Also from the range argument above, $x \geq 0$.

$$x + y + z = 9M - 24.8, \;\; M \in \mathbb{Z}_{>0}, \;\; x \geq 0$$

💡 The mean condition fixes the decimal part of $x+y+z$ to $.2$ — a one-line filter on the sum.

#7 Identify Subproblems 6.EE.B.5 Step 2

Split on $(\min, \max)$ of the full list.
Four cases, but $\min = 1$ AND $\max = 7$ gives range $6 \neq 7$, so three live cases: (I) $\min = 1$, $\max = z$ ($z > 7$, $x \geq 1$, so $z = 8$); (II) $\min = x$, $\max = 7$ ($x < 1$, $z \leq 7$, so $x = 0$); (III) $\min = x$, $\max = z$ ($0 \leq x < 1$, $z > 7$, $z = x + 7$).

$$\text{Case I: } z=8;\;\; \text{Case II: } x=0;\;\; \text{Case III: } z=x+7,\; 0\leq x<1$$

💡 Range $= 7$ has only three structural ways to happen — list them and treat each cleanly.

#2 Make a Systematic List 6.SP.B.5 Step 3

Case I: $z = 8$, $1 \leq x \leq y \leq 8$.
The sum $x + y + z = x + y + 8 \geq 10$, so $x + y + 8 \in \{11.2, 20.2\}$ (the only feasible mean values in range; $29.2$ would force $x+y > 21$, impossible with $y \leq 8$).
Sub-case $x + y = 3.2$ forces $1 \leq x \leq 1.6$, but then the sorted list begins $1, x, y, 2.2, 3.2, \dots$ so the median is $3.2$ — not integer.
Discard.
Sub-case $x + y = 12.2$ with $1 \leq x \leq y \leq 8$ gives $4.2 \leq x \leq 6.1$.

$$\text{Case I sub-case: } x+y=12.2, \;\; 4.2\leq x \leq 6.1$$

💡 Walk both feasible sums; one is killed by the median, the other narrows $x$ to a short interval.

#3 Eliminate Possibilities 6.SP.B.5 Step 4

Continue Case I.
The sorted list begins $1, 2.2, 3.2, \ldots$.
The 4th term is $\min(x, 5.2)$, the 5th (median) follows.
If $4.2 \leq x \leq 5.2$ the median is $5.2$ (not integer).
If $5.2 < x \leq 6.1$ the 5th term is $x$, so $x$ itself must be a positive integer in $(5.2, 6.1]$: only $x = 6$.
Then $y = 12.2 - 6 = 6.2$.
Triple: $(6, 6.2, 8)$.
Verify: sorted $\{1, 2.2, 3.2, 5.2, 6, 6.2, 6.2, 7, 8\}$, median $6$, mean $45/9 = 5$, range $7$.
All four conditions hold.

$(x, y, z) = (6,\, 6.2,\, 8)$ — valid

💡 Push the median condition through the sorted list; in this sub-case only $x = 6$ survives.

#3 Eliminate Possibilities 6.SP.B.5 Step 5

Case II: $x = 0$, $0 \leq y \leq z \leq 7$.
The sum $y + z \in \{2.2, 11.2\}$ (higher values impossible since $y, z \leq 7$).
The sub-case $y + z = 2.2$ leaves the sorted list beginning $0, y, 1, \dots$ with median $2.2$ — kill.
The sub-case $y + z = 11.2$ gives $4.2 \leq y \leq 5.6$.
The sorted list begins $0, 1, 2.2, 3.2, \dots$, and the 5th term is $\min(y, 5.2)$.
For it to be an integer with $y \geq 4.2$, only $y = 5$ works.
Then $z = 6.2$.
Triple: $(0, 5, 6.2)$.
Verify: sorted $\{0, 1, 2.2, 3.2, 5, 5.2, 6.2, 6.2, 7\}$, median $5$, mean $36/9 = 4$, range $7$.
Valid.

$(x, y, z) = (0,\, 5,\, 6.2)$ — valid

💡 Same machinery as Case I — the median equation forces $y$ to be an integer in a narrow window.

#2 Make a Systematic List 6.SP.B.5 Step 6

Case III: $z = x + 7$, $0 \leq x < 1$.
The sorted list begins $x, 1, 2.2, 3.2, \dots$, so the 5th term (median) is $\min(y, 5.2)$ provided $y > 3.2$ (otherwise the median is $3.2$, not integer).
So $y$ must be an integer in $(3.2, 5.2]$ — meaning $y \in \{4, 5\}$.
Mean condition: $24.8 + 2x + y + 7 = 9M$, i.e.
$31.8 + 2x + y = 9M$.
If $y = 4$: $35.8 + 2x = 9M$, only $M = 4$ works ($x = 0.1$).
If $y = 5$: $36.8 + 2x = 9M$, $M = 4$ needs $x = -0.4 < 0$, $M = 5$ needs $x = 4.1 \not< 1$ — both fail.

$$y=4 \Rightarrow x=0.1,\; z=7.1;\quad y=5 \Rightarrow \text{no solution}$$

💡 Two integer choices of $y$, each plugged into the mean equation — one yields a legal $x$, the other doesn't.

#3 Eliminate Possibilities 6.SP.B.5 Step 7

Verify the Case III candidate.
Triple: $(0.1, 4, 7.1)$.
Sorted: $\{0.1, 1, 2.2, 3.2, 4, 5.2, 6.2, 7, 7.1\}$, median $4$, mean $(24.8 + 11.2)/9 = 36/9 = 4$, range $7.1 - 0.1 = 7$.
All four conditions hold.
Total over all cases: $3$ valid triples.

$$(x, y, z) \in \big\{(6, 6.2, 8),\; (0, 5, 6.2),\; (0.1, 4, 7.1)\big\} \;\Rightarrow\; \textbf{(C)}$$

💡 Three cases, three survivors — the answer count drops cleanly out of the casework.

[1] #7 6.SP.B.5 Translate the conditions into algebra. Sum of the six fixed numbers: $1 + 2.2 +

[2] #7 6.EE.B.5 Split on $(\min, \max)$ of the full list. Four cases, but $\min = 1$ AND $\max =

[3] #2 6.SP.B.5 Case I: $z = 8$, $1 \leq x \leq y \leq 8$. The sum $x + y + z = x + y + 8 \geq 1

[4] #3 6.SP.B.5 Continue Case I. The sorted list begins $1, 2.2, 3.2, \ldots$. The 4th term is $

[5] #3 6.SP.B.5 Case II: $x = 0$, $0 \leq y \leq z \leq 7$. The sum $y + z \in \{2.2, 11.2\}$ (h

[6] #2 6.SP.B.5 Case III: $z = x + 7$, $0 \leq x < 1$. The sorted list begins $x, 1, 2.2, 3.2, \

[7] #3 6.SP.B.5 Verify the Case III candidate. Triple: $(0.1, 4, 7.1)$. Sorted: ${0.1, 1, 2.2,

Review

Reasonableness: The three triples each correspond to a different structural way the range $7$ can be realized — one with the new max at $z = 8$, one with the new min at $x = 0$, one with both endpoints fresh ($x = 0.1$, $z = 7.1$). That symmetry is a strong sign the casework is exhaustive. Also, each verified triple lands on a different mean ($5, 4, 4$) and a different median ($6, 5, 4$), reinforcing that they are genuinely distinct. The answer (C) $= 3$ matches.

Alternative: Tool #1 (Draw a Diagram) on a number line: plot the six fixed numbers, then think of $x \leq y \leq z$ as three sliders. The range condition forces at most one slider out past $7$ and at most one slider below $1$; the median (5th slot) is one of a handful of integer positions on the line. Visualizing the sorted slot of $y$ as either $\min(y, 5.2)$ or the lone integer immediately makes it obvious that only $y \in \{4, 5, 6\}$ can give an integer median, which is exactly the surface we walked algebraically. Same three triples emerge.

CCSS standards used (min grade 6)

6.SP.B.5 Summarize numerical data sets by reporting number of observations and measures (Translating mean, median, and range into algebraic conditions on $x + y + z$, the sorted 5th term, and $\max - \min$ — Grade 6 statistics summarization.)
6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Splitting on which of $\{x, 1\}$ is the min and which of $\{z, 7\}$ is the max, then solving each linear range equation.)

⭐ This AMC 10 problem only needs Grade 6 "mean / median / range" definitions plus careful case-by-case checking that you already know!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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