# Problem of the Week Problem D and Solution Two Birds

## Problem

Katya owns two cockatoos, an older white cockatoo and a younger Galah cockatoo. At present, the sum of the cockatoos’ ages is $$44$$ years. In $$n$$ years, where $$n >0$$, the white cockatoo’s age will be four times the Galah cockatoo’s age. If $$n$$ is an integer, determine the possible present ages of each cockatoo.

## Solution

Let $$g$$ represent the present age of the Galah cockatoo and $$w$$ represent the present age of the white cockatoo. Since the sum of their present ages is $$44$$, we have $$g+w=44$$ or $$w=44-g$$.

In $$n$$ years, the Galah cockatoo will be $$(g+n)$$ years old and the white cockatoo will be $$(44-g+n)$$ years old. At that time the white cockatoo will be four times older than the Galah cockatoo. Therefore, \begin{align} 4(g+n)&=44-g+n\\ 4g+4n&=44-g+n\\ 5g+3n&=44\\ g&=\frac{44-3n}{5} \end{align} We are looking for integer values of $$n$$ so that $$44-3n$$ is divisible by $$5$$.

When $$n=3$$, $$g=\frac{44-3n}{5}=\frac{44-3(3)}{5}=\frac{35}{5}=7$$. When $$g=7$$, $$w=44-g=44-7=37$$.

When $$n=8$$, $$g=\frac{44-3n}{5}=\frac{44-3(8)}{5}=\frac{20}{5}=4$$. When $$g=4$$, $$w=44-g=44-4=40$$.

When $$n=13$$, $$g=\frac{44-3n}{5}=\frac{44-3(13)}{5}=\frac{5}{5}=1$$. When $$g=1$$, $$w=44-g=44-1=43$$.

When $$n=18$$, $$g=\frac{44-3n}{5}=\frac{44-3(18)}{5}=\frac{-10}{5}=-2$$. Since $$g<0$$, $$n=16$$ does not produce a valid age for the Galah cockatoo. No higher value of $$n$$ would produce a value of $$g>0$$.

No integer values of $$n$$ between $$0$$ and $$18$$, other than $$3$$, $$8$$, and $$13$$, produce a multiple of $$5$$ when substituted into $$44-3n$$.

If today the white cockatoo is $$37$$ and the Galah cockatoo is $$7$$, then in $$3$$ years the white cockatoo will be $$40$$ and the Galah cockatoo will be $$10$$. The white cockatoo will be four times older than the Galah cockatoo since $$4\times 10=40$$.

If today the white cockatoo is $$40$$ and the Galah cockatoo is $$4$$, then in $$8$$ years the white cockatoo will be $$48$$ and the Galah cockatoo will be $$12$$. The white cockatoo will be four times older than the Galah cockatoo since $$4\times 12=48$$.

If today the white cockatoo is $$43$$ and the Galah cockatoo is $$1$$, then in $$13$$ years the white cockatoo will be $$56$$ and the Galah cockatoo will be $$14$$. The white cockatoo will be four times older than the Galah cockatoo since $$4\times 14=56$$.

Therefore, the possible present ages for the white cockatoo and Galah cockatoo are $$37$$ and $$7$$, or $$40$$ and $$4$$, or $$43$$ and $$1$$.