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2008 AMC 10B Problem 1

2008 AMC 10B Problem 1AMC 10 Step by Step

Name: Date: October 28, 2026

Basic Counting · difficulty 1 of 5 · about 1 min · position P1-10

A basketball player made $5$ baskets during a game. Each basket was worth either $2$ or $3$ points. How many different numbers could represent the total points scored by the player?

A) $2$
B) $3$
C) $4$
D) $5$
E) $6$

Answer

Key insight

Each time a 2-point basket becomes a 3-point basket, the total goes up by exactly 1. Six choices for the number of 3-pointers give six totals.

Solution

The only thing that changes the total is how many of the $5$ baskets are $3$ -pointers. So list every possibility.

The number of $3$ -pointers can be $0, 1, 2, 3, 4$ or $5$ . The rest are $2$ -pointers.
Work out each total:

$3$ -pointers	$2$ -pointers	Total
$0$	$5$	$0 + 10 = 10$
$1$	$4$	$3 + 8 = 11$
$2$	$3$	$6 + 6 = 12$
$3$	$2$	$9 + 4 = 13$
$4$	$1$	$12 + 2 = 14$
$5$	$0$	$15 + 0 = 15$

The six totals $10, 11, 12, 13, 14, 15$ are all different.

So there are $6$ possible totals.

The answer is $\boxed{\textbf{(E)}\ 6}$ .

Why this works

Trading one $2$ -point basket for a $3$ -point basket adds exactly $1$ point. So the totals climb by one at a time, from $10$ (all twos) to $15$ (all threes), and no total repeats. When a question asks "how many different totals", pick one count to vary and list the cases; remember to include both ends, zero and all.

The trap

Answering 5 because there are five baskets, and forgetting the cases where all baskets are 2-pointers or all are 3-pointers.

Concise solution

Let $k$ be the number of $3$ -point baskets, so the other $5 - k$ baskets are worth $2$ points each. The total is
$3k + 2(5 - k) = 10 + k.$
Since $k$ can be $0, 1, 2, 3, 4$ or $5$ , the possible totals are $10, 11, 12, 13, 14, 15$ : six different values, each achieved by exactly one $k$ .

The answer is $\boxed{\textbf{(E)}\ 6}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques
Introduction to Algebra, Chapter 7: Proportion (§7.4)

Techniques

Set up the equation/formula and compute; no special trick needed