2018 AMC 10A Problem 1

The expression looks scary, but it is the same small job repeated three times.

1. What does the power $-1$ mean? Raising a number to the power $-1$ flips it upside down. For example, $3^{-1} = \tfrac13$ and $\left(\tfrac43\right)^{-1} = \tfrac34$.
2. Where do we start? Brackets are worked from the inside out, so begin with the innermost one, $(2+1)$.
3. Layer 1. $2 + 1 = 3$. Flip: $\tfrac13$. Add $1$: $\tfrac13 + 1 = \tfrac43$.
4. Layer 2. Flip $\tfrac43$ to get $\tfrac34$. Add $1$: $\tfrac34 + 1 = \tfrac74$.
5. Layer 3. Flip $\tfrac74$ to get $\tfrac47$. Add $1$: $\tfrac47 + 1 = \tfrac{11}{7}$.

There are no brackets left, so we are done.

The answer is $\boxed{\textbf{(B)}\ \frac{11}{7}}$.

Every layer does the same two things: flip the fraction, then add $1$. Keeping the result as one simple fraction after each layer means you only ever flip a fraction like $\tfrac43$, never a messy sum. Notice the pattern too: $\tfrac43 \to \tfrac74 \to \tfrac{11}{7}$, where each new top is the old top plus the old bottom.

Work from the inside out. Each step is "take the reciprocal, then add $1$."

• Innermost: $(2+1)^{-1} = \tfrac13$, and $\tfrac13 + 1 = \tfrac43$.
• Next layer: $\left(\tfrac43\right)^{-1} = \tfrac34$, and $\tfrac34 + 1 = \tfrac74$.
• Outer layer: $\left(\tfrac74\right)^{-1} = \tfrac47$, and $\tfrac47 + 1 = \tfrac{11}{7}$.

The value is $\boxed{\textbf{(B)}\ \frac{11}{7}}$.