Sum of Cubed Numbers Game
Euclid, Pythagoras, Ptolemy, and Hypatia are playing a game where they all have to think of a number, and then cube that number 20 times. Hypatia doesn't want to cube large numbers, so she chooses the number 1. Euclid thinks the same thing and also chooses the number 1. However, Pythagoras and Ptolemy don't think ahead and Pythagoras chooses 2 and Ptolemy chooses -2. After they finish cubing their numbers (Pythagoras and Ptolemy take a while), all four players write their final numbers on a piece of paper. What is the sum of the numbers they wrote on the piece of paper?
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
The number 1 to any power is always 1, so no matter how many times Hypatia and Euclid cube their numbers, they will both always have the number 1. Thus, Hypatia and Euclid both write down the number 1.
Now, lets take a look at Pythagoras. He keeps cubing his result 20 times starting with the number 2. The numbers he has are \begin{align*}
2^3&\underset{\swarrow}{=}8 \\
8^3&\underset{\swarrow}{=}512 \\
512^3&\underset{\swarrow}{=}134217728 \\
\vdots
\end{align*}We can see that these numbers get out of hand pretty quickly (no wonder it took them so long), so instead, we look at the sum of the results because that is what we are trying to find. Recall that $ (-a)^3=-a^3 $. Because the number Ptolemy chose is the negation of what Pythagoras chose, after a certain number of cubes, whatever Pythagoras has, Ptolemy will have the opposite of that. Thus, whatever gigantic number Pythagoras has at the end of the game, Ptolemy will have the negation of that number, and they will add to zero.
Thus, our desired sum is \begin{align*}1+1&+(\text{some huge number}) + (-\text{the same huge number})\\
&=1+1+0=\boxed{2}.\end{align*}
Freeflows