AMC 8 · 2012 · #11
Grade 6 arithmeticProblem
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.
AMC 8 2012 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: 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 $x$.
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) $12$
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
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 $6$.
💡 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.
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 $x = 42 - 31 = 11$.
💡 The mean is just (sum)/(count). Multiplying both sides by the count turns the average condition into a simple sum-to-$42$ subproblem.
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.
💡 Tool #3 (Eliminate) on a multiple-choice problem: knock out anything that fails a stated condition. Four choices die fast.
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.
💡 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: $\textbf{(D)}\ 11$.
6.SP.B.5 Subproblem 1: find the mode. The listed numbers $3, 4, 5, 6, 6, 7$ already conta 6.SP.B.5 Subproblem 2: use the mean. Since mean $=$ mode $= 6$ and there are $7$ numbers, 6.SP.B.5 Eliminate the other answer choices against the unique-mode rule. (A) $x=5$: now 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.5Summarize 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.7Solve 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.
⭐ Once you spot that the mode has to be $6$, the rest is a Grade 6 mean problem — sum equals count times average.
More like this
Same archetype — closest grade level first.