# A girl scout troop uses 14 flashlight batteries on a three-night camping trip. If they are planning a seven-night trip, how many batteries should they bring?

Aug 29, 2016

#### Answer:

$33$ batteries would be sufficient.

#### Explanation:

There are several possible interpretations of this problem, resulting in different answers:

Suppose the flashlights are only used in the dark. Then the period for which they are used for the longer camping trip would be $\frac{7}{3}$ times as long as for the shorter one. The battery requirement would be approximately $\frac{7}{3} \cdot 14 = \frac{98}{3} = 32. \overline{6}$ which would be rounded up to $\textcolor{b l u e}{33}$.

But suppose the troop consists of $14$ scouts each with a flashlight that takes an single battery and a battery lasts exactly $3$ nights. Then after $6$ nights, you would have $14$ scouts all wanting a fresh battery to cover the $7$th night. So you would need a total of $3$ batteries per scout to cover the whole trip, that is $\textcolor{b l u e}{42}$ batteries.

This requirement could be reduced by labelling the batteries and cycling them. For example, all $14$ scouts could start with fresh batteries for the first night. Then $5$ of the partially used batteries could be labelled "$2$" and placed to one side, with the donors receiving fresh batteries. After the second night a different set of $5$ partially used batteries could be labelled "$1$" and placed to one side, with the donors receiving fresh batteries.

For the fourth and fifth nights, the scouts whose batteries had run out would receive fresh ones. On the last two nights the partially used batteries could be used instead. This scheme would use $\textcolor{b l u e}{33}$ batteries.

Suppose the flashlights are only used during daylight (!), in order to practice signalling using morse code, with the scouts supposed to be sleeping at night. A three-night camping trip includes up to four periods of daylight, while a seven-night trip includes up to eight. So the battery requirement may be double for the longer trip, which would give us a figure of $2 \cdot 14 = \textcolor{b l u e}{28}$ batteries.