Day 18: Splitting an Enormous Cake

After the first step $\frac34$ of the cake is remaining.

After the second step $\frac34 \times \frac45$ is remaining.

So after the final step there is $\frac34 \times \frac45 \times \dots \frac{99}{100}$ cake remaining.

But in a beautiful revelation we find that the fractions cancel to give $\frac{3}{100}$ which leads to the answer $\boxed{3\%}$ as required.

Let the cake weigh X kgs (it has to :D ).

Our mathematician gives $\Large \frac{X}{4}$ cake to his friend. $\Large \frac{3X}{4}$ cake remains.

Now $\Large \frac{1}{5}$ of $\Large \frac{3X}{4}$ is given to his second friend. $\Large \frac{3X}{4}$ * $\Large \frac{4}{5}$ = $\Large \frac{3X}{5}$ of cake remains.

Again, $\Large \frac{1}{6}$ of $\Large \frac{3X}{5}$ is given to this third friend. $\Large \frac{3X}{5}$ * $\Large \frac{5}{6}$ = $\Large \frac{3X}{6}$ cake remains.

Total cake given to his friends can be written as : $\Large \frac{X}{4}$ + $\Large \frac{3X}{4}$ * $\Large \frac{1}{5}$ + $\Large \frac{3X}{5}$ * $\Large \frac{1}{6}$ +...+ $\Large \frac{3X}{99}$ * $\Large \frac{1}{100}$ = $\Large \frac{X}{4}$ + $\Large {3X}$ ( $\Large \frac{1}{4}$ - $\Large \frac{1}{5}$ + $\Large \frac{1}{5}$ - $\Large \frac{1}{6}$ +... - $\Large \frac{1}{99}$ + $\Large \frac{1}{99}$ - $\Large \frac{1}{100}$ )

Subtracting this number from X gives quantity of cake remaining, which comes out to be equal to $\Large \frac{12X}{400}$ = $\Large \frac{3X}{100}$