Weather Dependent Number Change
The number 4 is written on my whiteboard. Every time it rains, I multiply the number on the whiteboard by $ \frac{2}{3} $, erase the original number, and write the new number on the whiteboard. When it snows, I multiply the number on the whiteboard by $ \frac{3}{5} $, and I replace the original number with the new number. It has rained 5 times and snowed 4 times this month. At the end of the month, what number is on the whiteboard?
- 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
Since it has rained 5 times, I have multiplied by $ \frac{2}{3} $ a total of 5 times. This is the same as multiplying by $ \left(\frac{2}{3}\right)^5 $, by definition of the exponent. Similarly, I have multiplied by $ \frac{3}{5} $ four times, or multiplied by $ \left(\frac{3}{5}\right)^4 $.
Because I started with 4 on the whiteboard at the beginning of the month, the number on the board at the end of the month is $ \displaystyle 4\left(\frac{2}{3}\right)^5\left(\frac{3}{5}\right)^4 $.
We know that $ \left(\frac{a}{b}\right)^n=\frac{a^n}{b^n} $, so we have $$4\left(\frac{2}{3}\right)^5\left(\frac{3}{5}\right)^4=4\left(\frac{2^5}{3^5}\right)\left(\frac{3^4}{5^4}\right).$$We can make this calculation simpler by rewriting the expression and using the exponent law $ \frac{a^m}{a^n}=a^{m-n} $, as shown below: \begin{align*} 4\left(\frac{2^5}{3^5}\right)\left(\frac{3^4}{5^4}\right)&=\left(\frac{4\cdot2^5}{5^4}\right)\left(\frac{3^4}{3^5}\right) \\ &=\left(\frac{4\cdot2^5}{5^4}\right)\left(3^{-1}\right)=\left(\frac{4\cdot2^5}{5^4}\right)\left(\frac{1}{3}\right).\end{align*}
Now we evaluate the remaining expressions: $$\left(\frac{4\cdot2^5}{5^4}\right)\left(\frac{1}{3}\right)=\frac{4\cdot32}{625}\cdot\frac{1}{3}=\boxed{\frac{128}{1875}}.$$
Freeflows