Henry Ernest Dudeney/Modern Puzzles/190 - Packing Cigarettes/Solution
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Modern Puzzles by Henry Ernest Dudeney: $190$
- Packing Cigarettes
- A manufacturer sends out his cigarettes in boxes of $160$;
- they are packed in $8$ rows of $20$ each, and exactly fill the box.
- Could he, by packing differently, get more cigarettes than $160$ into the box?
- If so, what is the greatest number that he could add?
Solution
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.
$\blacksquare$
Sources
- 1926: Henry Ernest Dudeney: Modern Puzzles ... (previous) ... (next): Solutions: $190$. -- Packing Cigarettes
- 1968: Henry Ernest Dudeney: 536 Puzzles & Curious Problems ... (previous) ... (next): Answers: $326$. Packing Cigarettes