AMC 8 · 2012 · #11

Easy mode Grade 6

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Problem

Here is a list of seven positive whole numbers: $3, 4, 5, 6, 6, 7, x$ . The last one, $x$ , is unknown.

For this list:

the mean (the average) equals
the median (the middle number when the list is sorted) equals
the mode (the number that shows up the most, and only one number shows up the most).

All three of those values are the same.

What number is $x$ ?

Pick an answer.

(A)

$hspace{.05in}5$

(B)

$hspace{.05in}6$

(C)

$hspace{.05in}7$

(D)

$hspace{.05in}11$

(E)

$hspace{.05in}12$

View mode:

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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