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I've been reading the, 'The Man of Numbers' book by Keith Devlin about Leonardo of Pisa who wrote the historic book on arithmetic around 1202 CE.

One of the problems, on page 70, goes as follows:

Three merchants find a purse lying in the road. The first asserts that the discovery would make him twice as wealthy as the other two combined. The second claims his wealth would triple if he kept the purse, and the third claims his wealth would increase five fold. How much concurrency would each receive.

Here is what I have so far.

//x for merchant one
//y for merchant two
//z for merchant three
x + y + z = 3

//I'm thinking 2y for the other two merchants, twice as much for the third 4x
4x + 2y = 6

3x = 3

This is about as far as I get. I realize that for five fold, five would be involved as a coefficient next to a variable but I can't seem to get any further.

So am I on the right track having three variables to represent each merchant? Although, I'm thinking the '= 3' seems redundant. Perhaps each of the variables represents the value of concurrency each would get.

Any thoughts any one


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What motivates $x+y+x=3$? I recommend fixing the least wealthy merchant's wealth at $1$, and then using three variables for the second merchant's wealth, the third merchant's wealth, and the value of the purse. –  alex.jordan Nov 30 '12 at 4:05
well why is x+y+z = 3? it should rather e x+t = 2(y+z) where t is the wealth they have found. Now try to work ahead –  Bhargav Nov 30 '12 at 4:07
It's important to understand that even if you knew the wealth of the three merchants and the value of the purse, you could double (or triple, or scale by whatever) all of these values and the merchants' statements would still be true. So you won't be able to find their values. This is why I suggest setting one value to $1$. You'll then be able to solve for the ratios of wealth that that the merchants have among each other. –  alex.jordan Nov 30 '12 at 4:08
An answer, with some of the working, can be found at www-history.mcs.st-and.ac.uk/Biographies/Mahavira.html See also page 386 of The Crest of the Peacock: Non-European Roots of Mathematics (Third Edition) by George Gheverghese Joseph (I found it at Google Books). –  Gerry Myerson Nov 30 '12 at 4:32
I don't have Liber Abaci here. This is an old puzzle that originated in India, and was taken westwards. The ones I know are a little more symmetric. Do you mean (for the second merchant) that his/her (probably his) wealth would triple, or that it would be $3$ times the combined wealth of the others? Wealth would triple is the easier version. –  André Nicolas Nov 30 '12 at 4:37
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1 Answer

up vote 2 down vote accepted

Let $x$ be merchant $1$'s wealth before the purse, $y$ for merchant $2$ and $z$ for merchant $3$ and $d$ the contents of the purse. We are given

$d+x=2(y+z) \\ y+d=3y \\ z+d=5z$

Then $d=2y=4z$, so $2y+x=2y+2z, x=2z, d=2x$

The solution is $d=2y=2x=4z$ and scales with the purse. As we had four unknowns and three equations this should not surprise.

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You meant $d=2x$, not $4x$, otherwise the first equation do not holds. –  carlop Nov 30 '12 at 10:04
@carlop: I fixed an earlier error as well. Thanks –  Ross Millikan Nov 30 '12 at 14:42
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