Geometric progressions

Algebra Level 4

99 infinitely long lines of people approach a pile of 1000000 cakes.

In the first line of people, the first person eats $\frac{1}{3}$ of the cakes, the second person eats $\frac{1}{3}$ of what the first person of the first line ate, the third person eats $\frac{1}{3}$ of what the second person ate, and so on. After the first line has passed, the first person of the second line eats $\frac{1}{4}$ of the remaining cakes, and the second person eats $\frac{1}{4}$ of what the first person of the second line ate, and so on. After all the lines have passed, a greedy person eats 7982 cakes. How many cakes remain?

Note: people are allowed to eat fractions of cakes.

The answer is 2018.

1 solution

Guilherme Niedu
Jan 30, 2017

The $n^{th}$ line eats $\large \frac{1}{n+2}$ of what the previous person ate. So, the line $n^{th}$ will eat $\large \frac{1}{n+2} + \frac{1}{(n+2)^2} + \frac{1}{(n+2)^3} + ... = \frac{1}{n+1}$ of the cake.

The total of cakes after the 99 lines passed the total of cakes will be:

$\large 1000000 \cdot (1 - \frac{1}{2}) \cdot (1 - \frac{1}{3}) \cdot (1 - \frac{1}{4}) \cdot ... \cdot (1 - \frac{1}{100})$

$\large = 1000000 \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdot ... \cdot \cdot \frac{99}{100}$

Each denominator will cancel out with the following numerator, remaining only the first numerator and the last denominator, or:

$\large = 1000000 \cdot \frac{1}{100} = 10000$

Take out the $7982$ cakes and $\color{#3D99F6} \fbox{2018}$ cakes remain.

Geometric progressions

The answer is 2018.

1 solution

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