2004 AMC 10B Problem 3

P1-10~1 minPrint

Sequences & Series Start from the end state / desired conclusion and reverse

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?

Show solution without answering

Key insight

From practice 1 to practice 5 there are only four doublings. Undo them by halving 48 four times.

Solution

Count the steps first. Going from practice $1$ to practice $5$ means going $1 \to 2$ , $2 \to 3$ , $3 \to 4$ , $4 \to 5$ . That is $4$ doublings, not $5$ .
We know the end (48) and want the start, so work backwards. Each doubling is undone by halving:

Practice	5	4	3	2	1
Free throws	$48$	$24$	$12$	$6$	$3$

Check forward: $3, 6, 12, 24, 48$ . Each number is twice the one before, and the fifth one is $48$ .

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

Why this works

When you know the last value and the rule that got you there, run the rule in reverse. The number of steps between term $1$ and term $5$ is one less than $5$ , so be careful to count the steps, not the terms. The wrong choice (B) $6$ comes from stopping one halving too early.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(A)}\ 3}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.4)
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)