AMC 8 · 2001 · #21
Grade 6 arithmeticnumber-theoryProblem
The mean of a set of five different positive integers is 15. The median is 18. The maximum possible value of the largest of these five integers is
Pick an answer.
AMC 8 2001 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 set has five different positive integers. Their mean is $15$ and their median is $18$. How large can the biggest number in the set possibly be?
Givens: Five different positive integers; Mean of the five numbers is $15$; Median of the five numbers is $18$; Answer choices: (A) $19$, (B) $24$, (C) $32$, (D) $35$, (E) $40$
Unknowns: The maximum possible value of the largest number
Understand
Restated: A set has five different positive integers. Their mean is $15$ and their median is $18$. How large can the biggest number in the set possibly be?
Givens: Five different positive integers; Mean of the five numbers is $15$; Median of the five numbers is $18$; Answer choices: (A) $19$, (B) $24$, (C) $32$, (D) $35$, (E) $40$
Plan
Primary tool: #11 Work Backwards
Secondary: #2 Make a Systematic List
The sum of all five numbers is locked at $75$. So making the largest number as big as possible is the same as making the other four as small as possible — that is the Tool #11 (Work Backwards) move: start from the target ('largest as big as possible') and back into the smallest legal values for the other slots. Tool #2 (Make a Systematic List) lines up the five slots in increasing order so the median and 'distinct positive' constraints are easy to enforce one slot at a time.
Execute — Answer: D
6.SP.B.5 Step 1 - Convert mean to sum.
- The five numbers average to $15$, so they add to $5 \times 15 = 75$.
- This total stays fixed no matter how we choose the individual numbers.
💡 Mean $\times$ count $=$ total. Holding the total fixed turns 'maximize one slot' into 'minimize the others'.
6.SP.A.3 Step 2 - Lay out the five slots in increasing order $a_1 < a_2 < a_3 < a_4 < a_5$.
- With five numbers, the median is the middle value, so $a_3 = 18$.
- The two slots below the median must be strictly less than $18$; the two above must be strictly greater than $18$.
💡 Sorting first makes the median rule visible: the third number is just the middle of the sorted list.
6.EE.B.8 Step 3 - Minimize $a_1$ and $a_2$.
- They must be positive integers and distinct, both less than $18$.
- The two smallest positive integers are $1$ and $2$, and both are below $18$, so $a_1 = 1$ and $a_2 = 2$ is legal and the smallest possible.
💡 To shrink the two below-median slots as much as possible, pick the smallest two distinct positive integers.
6.EE.B.8 Step 4 - Minimize $a_4$.
- It must be a positive integer strictly greater than $18$ and strictly less than $a_5$.
- The smallest integer greater than $18$ is $19$, so set $a_4 = 19$.
💡 The slot right above the median has to clear $18$, and the smallest integer that does that is $19$.
6.EE.B.7 Step 5 - Compute $a_5$ from the total.
- The other four slots add to $1 + 2 + 18 + 19 = 40$, so $a_5 = 75 - 40 = 35$.
- Check the order: $1 < 2 < 18 < 19 < 35$, all distinct positive integers — valid.
💡 Once the other four are pinned to their minimums, the leftover from $75$ is exactly the largest possible value.
6.SP.B.5 Convert mean to sum. The five numbers average to $15$, so they add to $5 \times 6.SP.A.3 Lay out the five slots in increasing order $a_1 < a_2 < a_3 < a_4 < a_5$. With f 6.EE.B.8 Minimize $a_1$ and $a_2$. They must be positive integers and distinct, both less 6.EE.B.8 Minimize $a_4$. It must be a positive integer strictly greater than $18$ and str 6.EE.B.7 Compute $a_5$ from the total. The other four slots add to $1 + 2 + 18 + 19 = 40$ Review
Reasonableness: Stress-test by trying to push $a_5$ above $35$. To do that we would need the other four to add to less than $40$. The two below-median slots are already at their floor ($1 + 2 = 3$), and the median is fixed at $18$, so the only place to cut is $a_4$. But $a_4$ must be an integer strictly between $18$ and $a_5$, so $a_4 \geq 19$ — already minimum. No more room. So $35$ is genuinely the ceiling, matching (D). The other choices fail: (A) $19$ is way too small (it's just the minimum value of $a_4$, not $a_5$); (B) $24$ and (C) $32$ leave the budget unused (we could shift weight to $a_5$); (E) $40$ would force the other four to sum to $35$, but $1 + 2 + 18 + 19 = 40 > 35$ already, impossible.
Alternative: Tool #6 (Guess and Check) on the answer choices, largest first. Try (E) $a_5 = 40$: the remaining four must sum to $75 - 40 = 35$ and include $18$, so the other three sum to $17$. With two below $18$ (distinct positive) and one above $18$ but below $40$, the minimum of those three is $1 + 2 + 19 = 22 > 17$ — impossible, rule out (E). Try (D) $a_5 = 35$: remaining four sum to $40$, and $1 + 2 + 18 + 19 = 40$ works exactly — valid. So (D) is the largest achievable value.
CCSS standards used (min grade 6)
6.SP.A.3Recognize that a measure of center for a numerical data set summarizes all of its values with a single number (Reading 'median $= 18$' as 'the middle (3rd) value of the five-number sorted list is $18$,' which fixes $a_3$ and splits the other four into two-below / two-above.)6.SP.B.5Summarize numerical data sets, including reporting the number of observations and measures of center (Converting 'mean $= 15$' on five values into the fixed total $a_1 + \cdots + a_5 = 75$, the budget we then distribute.)6.EE.B.7Solve real-world and mathematical problems by writing and solving equations of the form $x + p = q$ and $px = q$ (Solving $1 + 2 + 18 + 19 + a_5 = 75$ for $a_5 = 35$ once the four minimums are locked in.)6.EE.B.8Write an inequality of the form $x > c$ or $x < c$ to represent a constraint, and recognize that such inequalities have infinitely many solutions (Capturing the median and distinct-positive rules as $a_1 < a_2 < 18 < a_4 < a_5$ and picking the smallest integers that satisfy each strict inequality.)
⭐ Fixed mean means fixed total. To make one number as big as possible, push the other four down to their smallest legal values. Here the floors are $1, 2, 18, 19$, leaving $75 - 40 = 35$ for the largest — answer (D).
⭐ Fixed mean means fixed total. To make one number as big as possible, push the other four down to their smallest legal values. Here the floors are $1, 2, 18, 19$, leaving $75 - 40 = 35$ for the largest — answer (D).
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