AMC 10 · 2018 · #14

Grade 6 arithmetic
Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Integrality / Divisibility Screen
mean-median-mode-rangeoptimization-counting extremal-constructionbound-inequality-then-enumerate ↑ Prerequisites: mean-median-mode-range
📏 Medium solution 💡 2 insights
Problem
A list contains 2018 positive integers. Among them one value is the unique mode, meaning it appears more often than any other value, and it appears exactly 10 times. Find the smallest possible number of distinct values in the list.

Pick an answer.

(A)
202
(B)
223
(C)
224
(D)
225
(E)
234

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Extreme Principle

Tool #14 (Extreme Principle): fewer distinct values means each value must carry more of the 2018 entries, so we push every non-mode value to the largest count it is allowed to have. The unique-mode rule sets that ceiling. Tool #4 (Introduce a Variable): name the number of non-mode values d and turn the words into a single inequality 9d ≥ 2008. Tool #11 (Work Backwards): start from the fixed total 2018, peel off the 10 slots the mode already uses, and see how many slots the other values must cover.

1STEP 1

Cap every other value at nine

Unique mode = 10, so no other value can reach 10; each non-mode value is capped at 9.

(any non-mode value's count) ≤ 9
2STEP 2

Remove the mode's ten slots

The mode already fills 10 slots, so the other values must cover 2008 of the 2018 positions.

2018 - 10 = 2008
3STEP 3

Turn it into one inequality

Let d be the number of non-mode values; each covers at most 9, so 9d ≥ 2008.

9d ≥ 2008
4STEP 4

Divide and read the remainder

Dividing, 2008 ÷ 9 = 223 remainder 1, so 223 values leave one entry short — round up to 224.

2008 ÷ 9 = 223 r 1 → d ≥ 224
5STEP 5

Add the mode back

Add the mode to those 224 non-mode values: 224 + 1 = 225 distinct values, and the mode's 10 still beats every 9.

224 + 1 = 225 → (D)
Answer
225
Check the build directly: 223 values × 9 = 2007, plus one value × 1 = 1, plus the mode × 10 = 10, total 2007 + 1 + 10 = 2018 — exactly right, and the mode's 10 beats every other count of at most 9, so it is genuinely the unique mode. You cannot do better: with only 224 distinct values total (223 non-mode), those non-mode values cover at most 9 · 223 = 2007 entries, leaving 2018 - 10 - 2007 = 1 entry with nowhere to go. So 225 is forced, matching (D). The smaller choices (A) 202, (B) 223, (C) 224 each leave too few values to cover 2008 entries at 9 apiece.
💡Key takeaway

A unique mode of 10 caps everyone else at 9, so cover the other 2008 entries in groups of 9 — that needs 224 values, plus the mode makes 225.

  • Cap every other value at nine
  • Remove the mode's ten slots
  • Turn it into one inequality
  • Divide and read the remainder
  • Add the mode back

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