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2008 AMC 10B Problem 1

2008 AMC 10B Problem 1AMC 10 Step by Step

Name: Date: September 13, 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

With k three-point baskets the total is 10 + k, and k can be any of 0 through 5, so there are six totals.

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 $\boxed{\textbf{(E)}\ 6}$ .

Why this works

Every swap of a $2$ -pointer for a $3$ -pointer raises the total by exactly $1$ , so the totals form a run of consecutive integers with one value per choice of $k$ . Parametrizing by one count ( $k$ ) instead of tracking both kinds is the standard way to make such a "how many totals" question a one-line count.

The trap

Counting only the totals with at least one basket of each kind, or answering 5 because there are five baskets.

Common mistakes

Counting only the totals with at least one basket of each kind, or answering 5 because there are five baskets.
Listing $10$ through $15$ and counting the gaps ( $15 - 10 = 5$ ) instead of the integers themselves.

Techniques

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