AMC 8 · 2012 · #11

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

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

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.

Pick an answer.

(A)

$hspace{.05in}5$

(B)

$hspace{.05in}6$

(C)

$hspace{.05in}7$

(D)

$hspace{.05in}11$

(E)

$hspace{.05in}12$

AMC 8 2012 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Eliminate Possibilities

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.

1STEP 1

Only 6 already repeats (twice) while every other value appears once, so whatever x turns out to be, the unique mode is 6.

mode = 6

2STEP 2

The mean is 6 too, so the seven numbers total 7 × 6 = 42; subtract the six knowns (31) to get x = 11.

(3 + 4 + 5 + 6 + 6 + 7 + x)/7 = 6 → 31 + x = 42 → x = 11

3STEP 3

Check each choice: 5 and 7 tie 6 (mode not unique), 6 and 12 miss the mean — only x = 11 survives.

Sums: 5 → 36, 6 → 37, 7 → 38, 11 → 42, 12 → 43; need 42

4STEP 4

With x = 11 the sorted list is 3, 4, 5, 6, 6, 7, 11, whose middle value is 6 — median matches the mean and mode.

sorted: 3, 4, 5, 6, 6, 7, 11 → median = 6

Answer

hspace{.05in}11

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

\frac{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

\frac{31}{6}

≈ 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.

💡Key takeaway

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

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