(related to Problem: The Five Brigands)

The sum of $200$ doubloons might have been held by the five brigands in any one of $6,627$ different ways. Alfonso may have held any number from $1$ to $11.$ If he held $1$ doubloon, there are $1,005$ different ways of distributing the remainder; if he held $2,$ there are $985$ ways; if $3,$ there are $977$ ways; if $4,$ there are $903$ ways; if $5$ doubloons, $832$ ways; if $6$ doubloons, $704$ ways; if $7$ doubloons, $570$ ways; if $8$ doubloons, $388$ ways; if $9$ doubloons, $200$ ways; if $10$ doubloons, $60$ ways; and if Alfonso held $11$ doubloons, the remainder could be distributed in $3$ different ways. More than $11$ doubloons he could not possibly have had. It will scarcely be expected that I shall give all these $6,627$ ways at length. What I propose to do is to enable the reader, if he should feel so disposed, to write out all the answers where Alfonso has one and the same amount. Let us take the cases where Alfonso has $6$ doubloons, and see how we may obtain all the $704$ different ways indicated above. Here are two tables that will serve as keys to all these answers:—

Table I
$A = 6$
$B = n$
$C = (63 - 5n) + m$
$D = (128 + 4n) - 4m$
$E = 3 + 3m$
Table II
$A = 6$
$B = n$
$C = 1 + m$
$D = (376 - 16n) - 4m$
$E = (15n - 183) + 3m$

In the first table, we may substitute for $n$ any whole number from $1$ to $12$ inclusive, and m may be naught or any whole number from $1$ to $(31 + n)$ inclusive. In the second table, $n$ may have the value of any whole number from $13$ to $23$ inclusive, and $m$ may be naught or any whole number from $1$ to $(93 - 4n)$ inclusive. The first table thus gives $(32 + n)$ answers for every value of $n$; and the second table gives $(94 - 4n)$ answers for every value of $n.$ The former, therefore, produces $462$ and the latter $242$ answers, which together make $704,$ as already stated.

Let us take Table I., and say $n = 5$ and $m = 2;$ also in Table II. take $n = 13$ and $m = 0.$ Then we at once get these two answers:— $$\begin{array}{cr} &\text{Table I}\\ A =& 6\\ B =& 5\\ C =& 40\\ D =& 140\\ E =& 9\\ \hline &200\\ &\text{ doubloons} \end{array}$$

$$\begin{array}{cr} &\text{Table II}\\ A =& 6\\ B =& 13\\ C =& 1\\ D =& 168\\ E =& 12\\ \hline &200\\ &\text{ doubloons} \end{array}$$

These will be found to work correctly. All the rest of the $704$ answers, where Alfonso always holds six doubloons, may be obtained in this way from the two tables by substituting the different numbers for the letters $m$ and $n.$

Put in another way, for every holding of Alfonso the number of answers is the sum of two arithmetical progressions, the common difference in one case being $1$ and in the other $-4.$ Thus in the case where Alfonso holds $6$ doubloons one progression is $33 + 34 + 35 + 36 + \ldots + 43 + 44,$ and the other $42 + 38 + 34 + 30 + \ldots + 6 + 2.$ The sum of the first series is $462,$ and of the second $242$ —results which again agree with the figures already given. The problem may be said to consist in finding the first and last terms of these progressions. I should remark that where Alfonso holds $9,$ 10, or $11$ there is only one progression, of the second form.

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Project Gutenberg

  1. Dudeney, H. E.: "Amusements in Mathematics", The Authors' Club, 1917

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