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2020 AMC 12A Problem 13

Problem 13 of 25IntermediateAlgebra

There are integers a,a, b,b, and c,c, each greater than 1,1, such that NNNcba=N2536\sqrt[a]{N \sqrt[b]{N \sqrt[c]{N}}} = \sqrt[36]{N^{25}} for all N>1.N \gt 1. What is b?b?

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Solution

The left side equals NN raised to the exponent 1a+1ab+1abc,\dfrac{1}{a} + \dfrac{1}{ab} + \dfrac{1}{abc}, which must equal 2536.\dfrac{25}{36}. Because b,c≥2,b,c\ge2, this exponent is at most 74a.\dfrac{7}{4a}. If a≥3,a\ge3, it is at most 712<2536,\dfrac{7}{12}\lt\dfrac{25}{36}, so a=2.a=2. The equation becomes 12b+12bc=736.\dfrac{1}{2b}+\dfrac{1}{2bc}=\dfrac{7}{36}. Clearing denominators gives c(7b−18)=18.c(7b-18)=18. Set d=7b−18,d=7b-18, so dd is a positive divisor of 18.18. Among 1,2,3,6,9,18,1,2,3,6,9,18, only d=3d=3 makes b=d+187b=\dfrac{d+18}{7} an integer. Hence b=3b=3 and c=6.c=6. Thus, B is the correct answer.

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Concepts: radical · exponent

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