Henry Ernest Dudeney/Puzzles and Curious Problems/160 - Boxes of Cordite/Solution

From ProofWiki
Jump to navigation Jump to search

Puzzles and Curious Problems by Henry Ernest Dudeney: $160$

Boxes of Cordite
Cordite charges for $6$-inch howitzers were served out from ammunition dumps in boxes of $15$, $18$ and $20$.
"Why the three different sizes of boxes?" I asked the officer on the dump.
He answered: "So that we can give any battery the number of charges it needs without breaking a box.
This was an excellent system for the delivery of a large number of boxes,
but failed in small cases, like $5$, $10$, $25$ and $61$.
Now, what is the biggest number of charges that cannot be served out in whole boxes of $15$, $18$ and $20$?
It is not a very large number.


Solution

$97$


Proof

The dump officer gives boxes of $18$ until the remainder is a multiple of $5$.

Then, unless this is $5$, $10$ or $25$, the remainder is given in $15$s or $20$s.

The biggest number for which this breaks down is $72 + 25 = 97$.


Take the case of a higher numbers, such as $133$.

$6$ boxes of $18$ makes $108$, leaving $25$.

So you give one box of $18$, leaving $115$, which can be delivered in one box of $15$ and five of $20$.

But in the case of $97$, $72$ is the first and only case leaving a multiple of $5$, that is, $25$.

$\blacksquare$


Sources