2017 AMC 10A Problem 1

P1-10~1 minPrint

Sequences & Series Exponents, Logarithms & Radicals Set up the equation/formula and compute; no special trick needed Compute small cases, spot the pattern, generalize

What is the value of $(2(2(2(2(2(2+1)+1)+1)+1)+1)+1)$ ?

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Key insight

Start with the innermost parentheses and work outward. Each layer doubles the number and adds 1, giving 3, 7, 15, 31, 63, 127.

Solution

The only way into a stack of parentheses is from the inside, so start with the innermost pair and work outward.

The innermost part is $2 + 1 = 3$ .
Every layer after that has the same shape: take the number you have, double it, then add $1$ .
Apply that rule again and again, writing down each result:

Layer	Work	Result
1	$2 + 1$	$3$
2	$2 \cdot 3 + 1$	$7$
3	$2 \cdot 7 + 1$	$15$
4	$2 \cdot 15 + 1$	$31$
5	$2 \cdot 31 + 1$	$63$
6	$2 \cdot 63 + 1$	$127$

Check the count: the expression has six opening parentheses after the first one, one for each " $2(\ldots)+1$ ". So the process stops after layer $6$ .

The answer is $\boxed{\textbf{(C)}\ 127}$ .

Why this works

Nested parentheses always get worked from the inside out, one layer at a time. A table keeps you from losing your place. The results $3, 7, 15, 31, 63, 127$ are each one less than a power of $2$ ( $4, 8, 16, 32, 64, 128$ ), which is a handy check that no step went wrong.

Alternative approach

Use the pattern instead of multiplying. If a number is one less than a power of $2$ , then doubling it and adding $1$ gives one less than the next power of $2$ . For example, $7 = 8 - 1$ , and $2 \cdot 7 + 1 = 15 = 16 - 1$ . The first layer gives $3 = 4 - 1$ . Each of the five layers after it moves one power of $2$ up: $4 \to 8 \to 16 \to 32 \to 64 \to 128$ . So the answer is $128 - 1 = 127$ .

Concise solution

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

Work outward from the innermost parentheses. Each layer takes the previous result, doubles it, and adds $1$ :

$2+1 = 3,\quad 2\cdot 3+1 = 7,\quad 2\cdot 7+1 = 15,\quad 2\cdot 15+1 = 31,\quad 2\cdot 31+1 = 63,\quad 2\cdot 63+1 = 127.$

There are six " $2(\ldots)+1$ " layers, and the sixth one yields $127$ .

The answer is $\boxed{\textbf{(C)}\ 127}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 21: Sequences & Series (§21.1, §21.2, §21.3, §21.4)
Introduction to Algebra, Chapter 1: Follow the Rules (§1.6, §1.7, §1.8)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.1)
Intermediate Algebra, Chapter 17: More Sequences and Series (§17.1, §17.2)