Damn Monkey!

1259?

Nope.

99?

One answer for five men. Double it for ten. (No monkeys were hurt in this equation)

Let a be the number of coconuts to start with. After the first man and
the monkey took their coconuts, the number left was b = (4/5)*a - 1.
After the second man and the monkey took their coconuts, the number
left was c = (4/5)*b - 1. The third man left d = (4/5)*c - 1 coconuts.
The fourth man left e = (4/5)*d - 1 coconuts. The fifth man left
f = (4/5)*e - 1 coconuts. At the end, f = 5*g + 1, where g is the
number of coconuts each man got in the morning.

Using the last equation, substitute in the previous one to get

5*g + 1 = (4/5)*e - 1,
5*(5*g+2) = 4*e,
25*g + 10 = 4*e.

Now substitute that in the equation relating d and e:

25*g + 10 = 4*[(4/5)*d - 1],
5*(25*g+14) = 16*d,
125*g + 70 = 16*d.

Now substitute that in the equation relating c and d:

125*g + 70 = 16*[(4/5)*c - 1],
5*(125*g+86) = 64*c,
625*g + 430 = 64*c.

Now substitute that in the equation relating b and c:

625*g + 430 = 64*[(4/5)*b - 1],
5*(625*g+494) = 256*b,
3125*g + 2470 = 256*b.

Now substitute that in the equation relating a and b:

3125*g + 2470 = 256*[(4/5)*a - 1],
5*(3125*g+2726) = 1024*a,
15625*g + 13630 = 1024*a,
1024*a - 15625*g = 13630.

Any positive whole numbers a and g which are a solution of this
equation will give you a solution to your original problem. The
problem has been reduced to finding the value of a.

Notice that since 5 divides 15625 and 13630, but not 1024, that 5 must
divide a, so a = 5*h. Then

1024*5*h - 15625*g = 13630,

or, dividing by 5 throughout,

1024*h - 3125*g = 2726.

Similarly, since 2 divides 1024 and 2726, but not 3125, 2 must divide
g, so g = 2*i, and

1024*h - 3125*2*i = 2726,
512*h - 3125*i = 1363.

Now we can tell that i must be odd, so i = 2*j + 1, and

512*h - 3125*(2*j+1) = 1363,
512*h - 3125*2*j = 1363 + 3125 = 4488,
256*h - 3125*j = 2244.

Next, since 4 divides 256 and 2244, but 2 doesn't divide 3125, 4 must
divide j, so j = 4*k, and

256*h - 3125*4*k = 2244,
64*h - 3125*k = 561.

Similarly, k must be odd, so k = 2*m + 1, and

32*h - 3125*m = 1843.

Similarly, m must be odd, so m = 2*n + 1, and

16*h - 3125*n = 2484.

Now n must be divisible by 4, so n = 4*p, and

4*h - 3125*p = 621.

Again, p must be odd, so p = 2*q + 1, and

2*h - 3125*q = 1873.

Further, q must be odd, so q = 2*r+1, and

h - 3125*r = 2499.

One solution to this is r = 0, h = 2499. There are other, larger
ones, such as r = 1, h = 5624, and r = 100, h = 314999, but we will
stick with the smallest one. The general solution is h = 2499 + 3125*
r, r any nonnegative whole number.

Backtracking, we get q = 1, p = 3, n = 12, m = 25, k = 51, j = 204,
i = 409, h = 2499, g = 818, f = 4091, e = 5115, d = 6395, c = 7995,
b = 9995, and a = 12495. Sure enough.

Or, is it?

2,519!


By describing the numbers 2-10 in primes, you get
2=2
3=3
4=2x2
5=5
6=2x3
7=7
8=2x2x2
9=3x3
10=2x5

The smallest set of primes that has the ability to produce all of these numbers is (2,2,2,3,3,5,7), which multiplies out to 2520...thus the answer is 2520-1=2519.

Eos... my reasoning exactly, but I left out a factor of 2...?

oh sorry... I read the whole thing wrong... kind of sleepy...