2019 AMC 10A Problem 1

P1-10~1 minPrint

Exponents, Logarithms & Radicals Set up the equation/formula and compute; no special trick needed

What is the value of $2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9?$

Show solution without answering

Key insight

Work each tower from the top down. The first becomes 0 to the power 1, which is 0, then 2 to the power 0, which is 1.

Solution

A tower of exponents is worked out from the top down, so start with the smallest, innermost exponent in each term.

First term: $2^{\left(0^{\left(1^9\right)}\right)}$ .

Innermost: $1^9 = 1$ (one times itself any number of times is still one).
Next layer: $0^1 = 0$ .
Last layer: $2^0 = 1$ (any nonzero number to the power $0$ is $1$ ).

So the first term is $1$ .

Second term: $\left(\left(2^0\right)^1\right)^9$ .

Innermost: $2^0 = 1$ .
Then $1^1 = 1$ , and then $1^9 = 1$ .

So the second term is also $1$ .

Add: $1 + 1 = 2$ .

The answer is $\boxed{\textbf{(C)}\ 2}$ .

Why this works

The whole problem rests on two small rules that look alike: $0$ raised to a positive power is $0$ , while a nonzero number raised to the power $0$ is $1$ . The problem is built so that a rushed reader swaps them. Writing one layer per line, from the top of the tower down, keeps them straight.

Concise solution

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

Work each term from the innermost exponent outward.

First term: $1^9 = 1$ , so the exponent of $0$ is $1$ and $0^1 = 0$ . That leaves $2^0 = 1$ .

Second term: $2^0 = 1$ , then $1^1 = 1$ , then $1^9 = 1$ .

Adding, $1 + 1 = 2$ .

The answer is $\boxed{\textbf{(C)}\ 2}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 1: Follow the Rules (§1.6, §1.7, §1.8)
Introduction to Algebra, Chapter 11: Special Factorizations (§11.4)
Introduction to Algebra, Chapter 19: Exponents and Logarithms (§19.4)