AMC 10 · 2014 · #18
Grade 6 arithmeticA list of 11 positive integers has a mean of 10, a median of 9, and a unique mode of 8. What is the largest possible value of an integer in the list?
Pick an answer.
AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A list of $11$ positive integers has mean $10$, median $9$, and a unique mode of $8$. Find the largest value any single integer in the list can be.
Givens: The list has exactly $11$ positive integers; The mean is $10$, so the total of all $11$ numbers is $11 \times 10 = 110$; The median is $9$: when sorted, the $6$th number equals $9$; The mode is $8$ and it is unique: $8$ appears strictly more often than any other value; Answer choices: (A) $24$, (B) $30$, (C) $31$, (D) $33$, (E) $35$
Unknowns: The largest possible value of a single integer in the list
Understand
Restated: A list of $11$ positive integers has mean $10$, median $9$, and a unique mode of $8$. Find the largest value any single integer in the list can be.
Givens: The list has exactly $11$ positive integers; The mean is $10$, so the total of all $11$ numbers is $11 \times 10 = 110$; The median is $9$: when sorted, the $6$th number equals $9$; The mode is $8$ and it is unique: $8$ appears strictly more often than any other value; Answer choices: (A) $24$, (B) $30$, (C) $31$, (D) $33$, (E) $35$
Plan
Primary tool: #14 Extreme Principle
Secondary: #4 Introduce a Variable, #6 Guess and Check, #3 Eliminate Possibilities
The total of the eleven numbers is locked at $110$. To push one entry as high as possible (Tool #14, Extreme Principle), every other entry must be pushed as low as the rules allow. Tool #4 (Introduce a Variable) sorts the list as $a_1 \le \dots \le a_{11}$ so the median pins $a_6 = 9$ and the mode is forced into the low slots. The only real choice is how many $8$s to use: more $8$s raises the mode's count (letting other numbers repeat and stay small) but costs more total, so Tool #6 (Guess and Check) tests the count. Tool #3 (Eliminate Possibilities) then reads the outcome against the five answer choices — which, satisfyingly, are exactly the values the different $8$-counts produce.
Execute — Answer: E
6.SP.B.5 Step 1 Turn the three statistics into hard facts
- Decode each average.
- Mean $10$ over $11$ numbers means the total is $11 \times 10 = 110$.
- Median $9$ means that in the sorted list $a_1 \le a_2 \le \dots \le a_{11}$, the middle (6th) number is $a_6 = 9$.
- A unique mode of $8$ means $8$ shows up strictly more often than any other value.
- Since $8 < 9 = a_6$, every $8$ must live among the first five slots $a_1..a_5$.
💡 Mean, median, and mode are just three different summaries of the same list — each one is really a rule the numbers must obey.
6.EE.B.6 Step 2 Fix the total, shrink the rest
- The total is stuck at $110$.
- If one entry is to be huge, the other ten must be as small as the rules permit, because the largest entry equals $110$ minus the sum of the other ten.
- So the goal becomes: minimize $a_1 + \dots + a_{10}$ while still honoring $a_6 = 9$, the order, and the unique-mode rule.
💡 With a fixed pot of $110$, making one number big is the same as making all the others small.
6.SP.B.5 Step 3 Build the cheap lower half
- The mode $8$ sits in slots $a_1..a_5$.
- Using $8$ three times is the sweet spot: it lets a second value appear twice (needed later) without spending too much.
- Fill the other two low slots with the smallest positive integer, $1$.
- So the bottom five are $1, 1, 8, 8, 8$, which sums to $26$ and uses the mode three times.
💡 Three $8$s is enough to lead the vote, so the leftover slots can drop all the way to $1$.
6.NS.C.7 Step 4 Build the cheap upper half
- The middle is $a_6 = 9$.
- Slots $a_7..a_{10}$ must be $\ge 9$, and no value may reach $3$ copies (that would tie the mode's count of $3$).
- So $9$ may appear at most twice: use one more $9$ (now two total), then the smallest legal repeats above it, $10, 10$, then $11$.
- That gives $a_6..a_{10} = 9, 9, 10, 10, 11$, summing to $49$ with every value staying under three copies.
💡 Keep the numbers just above $9$, but never let any of them tie the three $8$s.
6.NS.B.3 Step 5 Read off the maximum
- The first ten numbers total $26 + 49 = 75$.
- The eleventh, and largest, is $110 - 75 = 35$.
- The full list $1, 1, 8, 8, 8, 9, 9, 10, 10, 11, 35$ checks out: total $110$ (mean $10$), $6$th value $9$ (median $9$), and $8$ appears three times while every other value appears at most twice (unique mode $8$).
- The largest possible integer is $35$, choice $\textbf{(E)}$.
💡 Everything below was squeezed to the legal minimum, so whatever is left over is forced into the last, biggest number.
6.SP.B.5 Decode each average. Mean $10$ over $11$ numbers means the total is $11 \times 1 6.EE.B.6 The total is stuck at $110$. If one entry is to be huge, the other ten must be a 6.SP.B.5 The mode $8$ sits in slots $a_1..a_5$. Using $8$ three times is the sweet spot: 6.NS.C.7 The middle is $a_6 = 9$. Slots $a_7..a_{10}$ must be $\ge 9$, and no value may r 6.NS.B.3 The first ten numbers total $26 + 49 = 75$. The eleventh, and largest, is $110 - Review
Reasonableness: Recheck the winning list $1, 1, 8, 8, 8, 9, 9, 10, 10, 11, 35$: sum $= 110$ so mean $= 10$; sorted, the $6$th term is $9$ so median $= 9$; counts are $8{:}3$, $9{:}2$, $10{:}2$, $1{:}2$, $11{:}1$, $35{:}1$, so $8$ is the one and only mode. All three conditions hold, and $35$ is the biggest listed choice, so nothing larger is on the table.
Alternative: Tool #3 (Eliminate Possibilities): test each possible number of $8$s. Five $8$s forces $8,8,8,8,8,9,9,9,9,10$ below, leaving $24$ (choice A). Four $8$s gives a top of $30$ (choice B). Two $8$s forces all other values distinct and gives $33$ (choice D). Three $8$s gives $35$ (choice E) — the largest. The wrong $8$-counts land exactly on the decoy answers, which is strong confirmation that three $8$s and $35$ is correct.
CCSS standards used (min grade 6)
6.SP.B.5Summarize numerical data sets by reporting number of observations and measures (Converting mean $10$ into total $110$, median $9$ into the sorted $6$th value, and the unique mode $8$ into a strict count condition.)6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Writing the largest entry as $110$ minus the sum of the other ten, so minimizing the rest maximizes it.)6.NS.C.7Understand ordering and absolute value of rational numbers (Ordering the list and choosing the smallest legal values in each slot without tying the mode's count.)6.NS.B.3Fluently add, subtract, multiply, and divide multi-digit decimals (Adding the ten small entries to $75$ and subtracting from $110$ to get the maximum $35$.)
⭐ When the total is fixed, make one number giant by squeezing every other number down to the smallest value the mean, median, and mode rules will allow.
⭐ When the total is fixed, make one number giant by squeezing every other number down to the smallest value the mean, median, and mode rules will allow.
More like this
Same archetype — closest grade level first.
