# Vessels received by the brother monarch

An eastern monarch sends 10.000 golden vessels to a brother monarch, whose kingdom is many days march distant. The gift is carried on camels. Each merchant who supplies camels for some part of the journey, demands as commissions 10% of what passes through his hands. Thus the first merchant hands over to the second 9.000 golden vessels. Altogether the vessels pass through 20 merchants. How many vessels does the brother monarch receive at the end? -- W.W Sawyer "Mathematician's Delight"

The time consuming solution is to manually calculate all the values, but there must be a logarithmic rule/formula that can used to find the result quickly for any number of merchants (without writing a computer program).

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What does this have to do with logarithms? – Rod Carvalho Sep 21 '12 at 20:17
@RodCarvalho It's included in the chapter of the book that is related with logarithms, so I guess that there must be a relation. – ierax Sep 21 '12 at 20:18
What happens when 10% of the vessels isn't a whole number? Can you divide vessels, do you round to the nearest integer, or do you always round up/down? – MartianInvader Sep 21 '12 at 20:24
@MartianInvader That's not inside the problem statement, so I assume it's fine to round as you wish :) – ierax Sep 21 '12 at 20:28
@Rod: Sawyer’s book was published in $1943$, when logarithms were by far the most practical way to evaluate $0.9^{20}$. – Brian M. Scott Sep 21 '12 at 21:20

Each time the vessels pass through a merchant, their number is multiplied by $.9$. Thus after $20$ merchants, the number of vessels is multiplied by $.9^{20} \approx 0.12158$. So in the end you have $10000(0.12157665459) \approx 1215.8$ vessels.

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I assume you need to do some kind of rounding at each step; I can't imagine that the vessels can be chopped into non-integer portions. – user22805 Sep 21 '12 at 20:20
There are a few more decimal places, so you should use $\approx$ signs. – Ross Millikan Sep 21 '12 at 20:21
@RossMillikan Good point, I've chopped off a bunch more decimal places and switched to $\approx$. – MartianInvader Sep 21 '12 at 21:00
As @David points out, the difficulties caused by unwillingness to chop vessels up need to be dealt with. But the problem didn't say whether to round always up, always down, or what. You expect the thieving middlemen to round up always, I suppose. – Lubin Sep 21 '12 at 21:47

Let $n_k$ be the number of vessels after passing through $k \geq 0$ merchants. We thus have

$$n_{k+1} = 0.9 n_k$$

with initial condition $n_0 = 10^4$. The general solution is thus $n_k = 0.9^k n_0$. If you want to take into account the fact that $n_k$ should be an integer, you can use the floor function $\lfloor \cdot \rfloor$, i.e.,

$$n_{k+1} = \lfloor 0.9 n_k \rfloor$$

Here's an experiment in Haskell (using the GHCi interpreter):

Prelude> let f n = floor (0.9 * fromIntegral n)
Prelude> let n0 = 10^4
Prelude> let ns = iterate f n0
Prelude> take 21 ns
[10000,9000,8100,7290,6561,5904,5313,4781,4302,3871,3483,3134,2820,2538,2284,2055,1849,1664,1497,1347,1212]


Hence, after $20$ merchants there will only be $1212$ vessels.

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It seems you need

$\begin{split} 10000*0.9^{20} &= 10000*(1-0.1)^{20} \\ &= 10000*(1 - 20*0.1 + \frac{20*19}{2} * 0.01 - \frac{20*19*18}{6}*0.001 + \frac{20*19*18*17}{24}*.0001 - \frac{20*19*18*17*16}{120}*.00001) \\ &\approx 10000*(1 - 2 + 1.9 - 1.14 + 0.4845 - 0.1550)\\ &= 1295 \end{split}$

feel free to add some terms if you need

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@RossMillikan I assumed he wanted a way to compute it by hand, of course given a calculator or a log table one should not hesitate :) – gt6989b Sep 21 '12 at 20:23
Why bother to expand the power? Why not just calculate $0.9^{20}?$ This is off by about $80$ – Ross Millikan Sep 21 '12 at 20:24

This isn’t so much an answer (though it does contain a good estimate) as a brief visit to the past.

Having grown up before calculators, and having memorized the $5$-place decimal logs of $2,3$, and $7$ over $50$ years ago, I was curious to see how close I could come using only easy hand computation. We want $10000(0.9)^{20}$, whose log is $4+20\log0.9\approx4+20(0.95424-1)=3.0848$, so it’s going to be a bit over $1000$.

$\log 5-\log 4\approx 0.69897-0.60206=0.09691$ and $\log6-\log5\approx0.77815-0.69897=0.07918$, so it’s between $1200$ and $1250$, closer to the former. After this it gets harder, but $1200=5\cdot240$, $1250=5\cdot250$, and by good fortune I recognize that interval as containing $243=3^5$, so we might try $3^5\cdot5=1215$: $\log1215=5\log3+\log5\approx 5\cdot0.47712+0.69897=3.08457$, so $1215$ is pretty close. (Hey, it beats counting sheep!)

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