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

by : $160$

 * 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, $25$.