2023 AMC 8

Complete problem set with solutions and individual problem pages

Problem 6 Easy

The digits 2,0,2, and 3 are placed in the expression below, one digit per box. What is the maximum possible value of the expression?

  • A.

    0

  • B.

    8

  • C.

    9

  • D.

    16

  • E.

    18

Answer:C

Solution 1

First, let us consider the case where 0 is a base: This would result in the entire expression being 0. Contrastingly, if 0 is an exponent, we will get a value greater than 0. 3^2\times2^0=9 is greater than 2^3\times2^0=8 and 2^2\times3^0=4. Therefore, the answer is \boxed{\textbf{(C) }9}.

 

Solution 2

The maximum possible value of using the digits 2,0,2, and 3: We can maximize our value by keeping the 3 and 2 together in one power (the biggest with the biggest and the smallest with the smallest). This shows 3^{2}\times2^{0}=9\times1=9. (We don't want 0^{2} because that is 0.) It is going to be \boxed{\textbf{(C)}\ 9}.

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