jmc

Question

The number 4 is written on my whiteboard. Every time it rains, I multiply the number on the whiteboard by 23, erase the original number, and write the new number on the whiteboard. When it snows, I multiply the number on the whiteboard by 35, 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?

Accepted Answer

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

(\frac{2}{3})^{5}

, by definition of the exponent. Similarly, I have multiplied by

\frac{3}{5}

four times, or multiplied by

(\frac{3}{5})^{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

4 (\frac{2}{3})^{5} (\frac{3}{5})^{4}

.

We know that

(\frac{a}{b})^{n} = \frac{a ^{n}}{b ^{n}}

, so we have

4 (\frac{2}{3})^{5} (\frac{3}{5})^{4} = 4 (\frac{2 ^{5}}{3 ^{5}}) (\frac{3 ^{4}}{5 ^{4}}) .

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:

4 (\frac{2 ^{5}}{3 ^{5}}) (\frac{3 ^{4}}{5 ^{4}}) = (\frac{4 \cdot 2 ^{5}}{5 ^{4}}) (\frac{3 ^{4}}{3 ^{5}}) = (\frac{4 \cdot 2 ^{5}}{5 ^{4}}) (3^{- 1}) = (\frac{4 \cdot 2 ^{5}}{5 ^{4}}) (\frac{1}{3}) .

Now we evaluate the remaining expressions:

(\frac{4 \cdot 2 ^{5}}{5 ^{4}}) (\frac{1}{3}) = \frac{4 \cdot 32}{625} \cdot \frac{1}{3} = \frac{128}{1875} .