2017 vs 2019! Who wins?

Algebra Level 2

Suppose that:

$\large A = (1 +2017)^{2017} \\ \\ \large B = (1 + 2019)^{2016}$

Then which of the following is true?

A > B none of them A < B A = B

1 solution

Naren Bhandari
Oct 15, 2017

$\begin{aligned} & A = (1+2017)^{2017} = 2018^{2017} \\ & B=(1+2019)^{2016} = 2020^{2016}\end{aligned}$

Let $p = \frac{A}{2018} = \frac{2018^{2017}}{2018} = 2018^{2016}$ . Then

$\begin{aligned}\frac{p}{B} = \frac{A}{2018B} & = \frac{2018^{2016}}{2020^{2016}} \\ & = 2018\left( \frac{2018}{2020}\right)^{2016} \\& = 2018\left(1-\frac{2}{2020}\right)^{2016 }\\ & = 2018q > 1 \end{aligned}$ .

where $q$ exists in $0<q <1$ Thus $A>B$ .

Very interesting approach to this question! :))

Syed Hamza Khalid - 3 years, 7 months ago

Why not just calculate A/B diectly....you will immediately have your next to the last expression... 2018(1 - 2/2020)^2016 >1 ?

Greg Grapsas - 2 years, 6 months ago

I doubt this solution and your result. At first, in the 2nd line of calculating p/B, you multiply by 2018 out of nothing. And you didnt proof that 2018*(2018/2020)^2016 is less than 1. Wolframalpha gives different result aswell, though this is no proof either. Can you give a more convincing solution?

Eric Scholz - 1 year, 10 months ago

2017 vs 2019! Who wins?

1 solution

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