2017 AMC 10A Problem 1

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 $\boxed{\textbf{(C)}\ 127}$.

Nested expressions are evaluated inside out, one layer at a time. Noticing that each output is one less than a power of $2$ ($3 = 2^2-1$, $7 = 2^3-1$, and so on) is a useful check: after $k$ layers the value is $2^{k+1}-1$, so six layers give $2^7 - 1 = 127$.

If $v$ is one less than a power of two, say $v = 2^m - 1$, then $2v + 1 = 2^{m+1} - 1$ is again one less than a power of two. Starting from $3 = 2^2 - 1$ and applying five more layers lands on $2^7 - 1 = 127$ without any multiplication.

• Miscounting the layers (there are six doublings) and stopping at 63 or continuing to 255.
• Adding the $1$ before doubling, which turns "$2v+1$" into "$2(v+1)$" and produces even numbers.