AMC 8 · 2012 · #11

Grade 6 arithmetic

Archetype Small-Domain Constrained Search

mean-median-mode-rangelinear-equations-one-var logical-deductionconvert-to-algebra ↑ Prerequisites: mean-median-mode-rangemulti-digit-arithmetic

📏 Medium solution 💡 3 insights

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Problem

The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and $x$ are all equal. What is the value of $x$ ?

Pick an answer.

(A)

$hspace{.05in}5$

(B)

$hspace{.05in}6$

(C)

$hspace{.05in}7$

(D)

$hspace{.05in}11$

(E)

$hspace{.05in}12$

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Toolkit + CCSS Solution

Understand

Restated: Seven positive integers are $3, 4, 5, 6, 6, 7$, and $x$. Their mean, median, and unique mode are all the same number. Find $x$.

Givens: The seven values are $3, 4, 5, 6, 6, 7, x$; Mean $=$ Median $=$ Mode; The mode is unique (exactly one number appears most often); All values are positive integers; Answer choices: (A) $5$, (B) $6$, (C) $7$, (D) $11$, (E) $12$

Unknowns: The value of $x$

Understand

Restated: Seven positive integers are $3, 4, 5, 6, 6, 7$, and $x$. Their mean, median, and unique mode are all the same number. Find $x$.

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #6 Guess and Check, #7 Identify Subproblems

Three conditions must all hold at once: mean, median, and unique mode are equal. The cleanest path is Tool #7 (Identify Subproblems) — pin down the mode first (it constrains itself), then use it to force the mean. Because $6$ already appears twice and no other number repeats, the mode is forced to be $6$ — that single observation collapses the problem. Tool #3 (Eliminate Possibilities) is a natural double-check: scan the five answer choices and drop any that would create a tied mode or shift the mean off $6$. Tool #6 (Guess and Check) backs up the arithmetic by plugging the winning $x$ back into the list.

Execute — Answer: D

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

Subproblem 1: find the mode.
The listed numbers $3, 4, 5, 6, 6, 7$ already contain $6$ twice and every other value once.
For the mode to be unique, $x$ must not turn another number into a tie with $6$.
So either $x = 6$ (making $6$ appear three times) or $x$ is a brand-new value (keeping $6$'s count at two while every other count stays at one).
Either way, the unique mode is $6$.

$$\text{mode} = 6$$

💡 Splitting the three conditions and tackling the mode first is the Tool #7 move — one subproblem locks in a number we can use everywhere else.

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

Subproblem 2: use the mean.
Since mean $=$ mode $= 6$ and there are $7$ numbers, the sum of all seven values must be $7 \times 6 = 42$.
Add the six known values: $3 + 4 + 5 + 6 + 6 + 7 = 31$.
So $x = 42 - 31 = 11$.

$$\dfrac{3 + 4 + 5 + 6 + 6 + 7 + x}{7} = 6 \;\Rightarrow\; 31 + x = 42 \;\Rightarrow\; x = 11$$

💡 The mean is just (sum)/(count). Multiplying both sides by the count turns the average condition into a simple sum-to-$42$ subproblem.

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

Eliminate the other answer choices against the unique-mode rule.
(A) $x=5$: now $5$ appears twice, tying $6$ — mode not unique.
(B) $x=6$: list becomes $3,4,5,6,6,6,7$, mode is $6$ (unique), but mean $= 37/7 \ne 6$.
(C) $x=7$: $7$ ties $6$ — mode not unique.
(E) $x=12$: mode is $6$ (unique), but mean $= 43/7 \ne 6$.
Only (D) $x = 11$ survives.

$$\text{Sums: } 5 \to 36,\; 6 \to 37,\; 7 \to 38,\; 11 \to 42,\; 12 \to 43;\; \text{need } 42$$

💡 Tool #3 (Eliminate) on a multiple-choice problem: knock out anything that fails a stated condition. Four choices die fast.

#6 Guess and Check 6.SP.B.5 Step 4

Verify the median.
With $x = 11$ the sorted list is $3, 4, 5, 6, 6, 7, 11$.
The middle (4th) value is $6$, matching the mean and the mode.
All three conditions hold.

$$\text{sorted: } 3, 4, 5, \mathbf{6}, 6, 7, 11 \;\Rightarrow\; \text{median} = 6$$

💡 Tool #6 (Guess and Check) finishes the job — plug the candidate back into the original setup to make sure every requirement (not just the one you used) is satisfied. Answer: $\textbf{(D)}\ 11$.

[1] #7 6.SP.B.5 Subproblem 1: find the mode. The listed numbers $3, 4, 5, 6, 6, 7$ already conta

[2] #7 6.SP.B.5 Subproblem 2: use the mean. Since mean $=$ mode $= 6$ and there are $7$ numbers,

[3] #3 6.SP.B.5 Eliminate the other answer choices against the unique-mode rule. (A) $x=5$: now

[4] #6 6.SP.B.5 Verify the median. With $x = 11$ the sorted list is $3, 4, 5, 6, 6, 7, 11$. The

Review

Reasonableness: All three statistics land on $6$: the mode is $6$ (appears twice, more than any other value), the sorted middle is $6$, and the mean $42/7 = 6$. The chosen $x = 11$ is larger than every other value, which is fine — it just sits at the end of the sorted list and pulls the sum up to exactly $42$. Sanity check: the six known numbers average $31/6 \approx 5.17$, which is below $6$; we need the seventh number to drag the average up to $6$, so $x$ must be bigger than $6$ — that immediately rules out (A), (B), (C) without any arithmetic.

Alternative: Tool #6 (Guess and Check) on the answer choices alone: for each candidate $x$, compute the sum and divide by $7$. (A) $36/7$, (B) $37/7$, (C) $38/7$, (D) $42/7 = 6$, (E) $43/7$. Only (D) gives an integer mean equal to $6$. This skips the mode reasoning entirely and lands on the answer in three lines — a classic AMC time-saver.

CCSS standards used (min grade 6)

6.SP.B.5 Summarize numerical data sets in relation to their context, including measures of center (mean, median, mode) (Identifying that the unique mode is $6$ from the repeated value, computing the mean as sum-over-count, and verifying the median of the sorted list — all three measures of center on the same data set.)
6.EE.B.7 Solve real-world and mathematical problems by writing and solving equations of the form $x + p = q$ (Turning the mean condition into $31 + x = 42$ and solving for $x = 11$.)

⭐ Once you spot that the mode has to be $6$, the rest is a Grade 6 mean problem — sum equals count times average.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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