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2018 AMC 10A Problem 1

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2018 AMC 10A Problem 1AMC 10 Step by Step

Name: Date: September 13, 2026

Exponents, Logarithms & Radicals · difficulty 1 of 5 · about 1 min · position P1-10

What is the value of $\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1?$

A)
$\frac58$
B)
$\frac{11}7$
C)
$\frac85$
D)
$\frac{18}{11}$
E)
$\frac{15}8$

Answer

Key insight

A negative-one exponent just flips the fraction; work from the innermost parentheses outward, flipping and adding 1 three times.

Solution

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

Why this works

An exponent of $-1$ means reciprocal, so $\left(\tfrac pq\right)^{-1} = \tfrac qp$ . Nested expressions of this kind are evaluated strictly from the innermost parentheses outward; keeping each intermediate result as a single fraction prevents errors. Notice the pattern $\tfrac{p}{q} \to \tfrac{p+q}{p}$ at every layer, a miniature continued fraction.

The trap

Flipping a fraction after adding 1 to the wrong layer, or treating the exponent -1 as a minus sign rather than a reciprocal.

Common mistakes

Flipping a fraction after adding 1 to the wrong layer, or treating the exponent -1 as a minus sign rather than a reciprocal.
Stopping one layer early at $\tfrac74$ or flipping the final answer to $\tfrac{7}{11}$ .

Techniques

Set up the equation/formula and compute; no special trick needed