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2018 AMC 10B Problem 14

2018 AMC 10B Problem 14AMC 10 Step by Step

Name: Date: September 13, 2026

Statistics & Data · difficulty 3 of 5 · about 3 min · position P11-15

A list of $2018$ positive integers has a unique mode, which occurs exactly $10$ times. What is the least number of distinct values that can occur in the list?

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

Answer

Key insight

Every non-mode value appears at most 9 times, so the other 2008 entries need at least ceil(2008/9) = 224 values, plus the mode makes 225.

Solution

The mode occupies $10$ of the $2018$ entries, leaving $2008$ entries to be filled by other values. Because the mode is unique, every other value appears at most $9$ times.

To use as few distinct values as possible, make each of the other values appear as often as allowed, $9$ times. Then the number of other values $k$ must satisfy $9k \ge 2008$ , so
$k \ge \frac{2008}{9} = 223.1\ldots, \qquad k \ge 224.$
This is achievable: take $223$ values appearing $9$ times each ( $2007$ entries) and one more value appearing once.

Including the mode, the minimum number of distinct values is $224 + 1 = 225$ .

The answer is $\boxed{\textbf{(D)}\ 225}$ .

Why this works

Minimizing the number of distinct values means packing the non-mode entries as tightly as the "unique mode" rule permits, which caps every other frequency at $9$ . The bound is a ceiling, and the construction shows it is attained. Always finish by counting all the values asked for, here the mode included.

The trap

Forgetting to add the mode itself (answering 224), or dividing 2008 by 9 and rounding down to 223.

Common mistakes

Forgetting to add the mode itself (answering 224), or dividing 2008 by 9 and rounding down to 223.
Allowing other values to appear $10$ times as well, which would destroy the uniqueness of the mode and give $202$ .

Techniques

Bound the quantity above/below or estimate to pin it down · Consider the largest/smallest element or boundary case