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2004 AMC 10B Problem 3

2004 AMC 10B Problem 3AMC 10 Step by Step

Name: Date: September 13, 2026

Sequences & Series · difficulty 1 of 5 · about 1 min · position P1-10

At each basketball practice last week, Jenny made twice as many free throws as she made at the previous practice. At her fifth practice she made $48$ free throws. How many free throws did she make at the first practice?

A)
$3$
B)
$6$
C)
$9$
D)
$12$
E)
$15$

Answer

Key insight

Doubling four times separates practice 1 from practice 5, so the first count is 48 divided by 2^4.

Solution

Going from practice $1$ to practice $5$ involves four doublings (from $1$ to $2$ , $2$ to $3$ , $3$ to $4$ , and $4$ to $5$ ). Reversing the process, undo each doubling by halving:
$48 \to 24 \to 12 \to 6 \to 3.$

Equivalently, if the first practice had $x$ free throws, then the fifth had $2^4 x = 16x = 48$ , so $x = 3$ .

The answer is $\boxed{\textbf{(A)}\ 3}$ .

Why this works

A quantity that doubles each step is a geometric sequence, $a_n = a_1 \cdot 2^{\,n-1}$ ; the exponent is the number of steps between the terms, one less than the term number. When the late term is known, working backwards by repeated halving is the quickest and least error-prone route.

The trap

Halving one time too few (answering 6) or one time too many, because 'fifth practice' means four doublings, not five.

Common mistakes

Halving one time too few (answering 6) or one time too many, because 'fifth practice' means four doublings, not five.
Dividing $48$ by $5$ or by $2 \cdot 5$ as if the growth were additive rather than multiplicative.

Techniques

Start from the end state / desired conclusion and reverse