Henry Ernest Dudeney/Modern Puzzles/190 - Packing Cigarettes/Solution

by : $190$

 * Packing Cigarettes

Solution

 * Dudeney-Modern-Puzzles-190-solution.png

Let the radius of the cross-section of a cigarette be $1$ unit.

It immediately follows that $160$ cigarettes when packed as on the left, in $8$ rows of $20$, is $8 \times 2 = 16$ units high.

When packed as on the right, however, there are $4$ rows of $20$ and $4$ rows of $19$, making $156$ cigarettes.

However, the centres of the cross-sections of the cigarettes in row $n$ are $1 + \paren {n - 1} \sqrt 3$ above the bottom of the pack.

Let us arrange a $9$th row of $20$ on top of that $8$th row.

The top of the top row of cigarettes is now $2 + 8 \sqrt 3 \approx 15.9$ units high, and definitely less than $16$ units.

So we can get $16$ more cigarettes in the box if we pack them in the second way.