What number, between 5 and 5000, are we able to divide with no decimals or remainder for the problem below?

Question

A rich collector of gold coins has left a will giving instructions stating how his coins are to be
distributed to his butler and family (5 children and 5 grandchildren) after his death. His instructions
are that, first of all, one coin is to be given to his butler. Then his eldest child is to get one fifth of the
remaining coins. Another coin is then to be given to his butler. Then his second eldest child is to get
one fifth of the remaining coins. This procedure is to be repeated until all five children have their
share. Finally, the remaining coins are to be divided evenly among the five grandchildren. You are
told that the collector had between 5 and 5000 coins.

How many gold coins, exactly, did the collector have in his collection?

Answer 1

The answer is that the collector had 3121 coins in his collection.

 It is straightforward to check that this is an answer.

To show that it is the only answer less than 5000 can be done by a tedious process.

It starts off by noting that a possible original number of coins must be one more than a multiple of 5.  This condition is required because after we pay the butler his first coin, we must have a multiple of 5 coins left if the first son is to be paid his 1/5 th as a integer number of coins.  So the original number of coins could be one more than:

4995, 4991, 4986, 4981, 4976, 4970  , ... , 3121  , ...

After we pay the butler then the first son,  this numbers become

3996 ,  3992, 3988,  3984, 3980, 3976 ....,  2496.  ...

Now we pay the butler again leaving

3995 , 3991, 3987, 3983, 3979, 3975 ..., 2495, ...

The next step is to pay son 2  his 1/5 th.  To do so with an integer number of coins, the number at this

point must be a multiple of 5.   The gap between numbers of the last sequence  with this property is 20

So we keep only this subset of the sequence and sequence becomes

3995, 3975, 3955, 3935, 3915, 3895,... ,  2495       Now we pay son 2 resulting in

3196 , 3180, 3164, 3184, 3132, 3116,  ... ,  1996

Now we pay the butler again   resulting in

3195, 3179, 3163, 3147 , 3131, ... , 1995

The next step is to pay son 3 his 1/5 th.   To do so with an integer number of coins, the number at this

point must be a multiple of 5. The gap between numbers of the last sequence with this property is 80.

So we keep only the subset with is property and the sequence becomes

3195, 3115, 3035, 2955, 2875, 2795,  ... , 1995

Now we pay son 3 his 1/5  resulting in

2556, 2492, 2428, 2364 , 2300, 2236,  ..., 1596

Now we pay the butler again,   resulting in

2555, 2491, 2427, 2363, 2299, 2235,.  ... , 1595

The next step is to pay son 4.   In order to do this with an integer number of coins, the number at this

point must be a multiple of 5.  The gap between numbers with this property is 320

So we keep only the subset with this property   leaving

2555, 2235, 1915, 1595, 1275,  and some lower numbers than will not work out

Now we pay son 4  his 1/5 leaving

2044, 1788, 1532, 1276, 1020,   and some lower numbers that will not work out

Now we pay the butler again   leaving

2043 , 1787, 1531, 1275, 1019,  and some smaller numbers

The only number left that is a multiple of 5 is 1275. 

Using this number, we pay the fifth son leaving 1020 .  This is a multiple of 5 so each grandchild

gets  204 - an integer.

If we trace back the 1275 from the last sequence, we see that it results from starting with 3121 coins.

 

Note: If we had to pay the butler after the fifth son (before paying the grandchildren) there would be no

solution at all.

Answer 2

This is a really fun problem! It reminds me of a Futurama episode, where the butler is to be paid a pittance, in 20 equal installment of 1/20th of a pittance each, while the kid who doesn't know the value of a dollar gets the rest of the $10 million fortune. Anyway, on to the math. It's quite lengthy, so bear with me. I also hope this is a class in which you've learned modular arithmetic, or else my explanation will stop making sense near the end.

 

Let's call the amount-to-be-divvied N, where we know that 4 < N < 5001. Let's say that a "round of giving" consists of giving the butler his coin, then giving away 1/5 of what's left. Let n_k denote the amount left over after k rounds of giving. We'll also say that n₀=N, since there are N coins left to be distributed after 0 rounds of giving.

 

Thus, we have

n₀ = N, while

n_k+1 = (4/5)(n_k-1). This expression is because the butler starts the (k+1)^st round by taking his coin, then the (k+1)^st child takes 1/5 of what's left (there's n_k-1 left at this point), leaving 4/5 of that amount remaining. This will happen for five rounds.

 

Thus, 

n₀ = N

n₁ = (4/5)(N-1)

n₂ = (4/5)[(4/5)(N-1)-1], etc. I'll let you write down exactly what n₅ is (mostly because there are 5 levels of nested parenthesis, and it will be essentially unreadable without being able to alter parenthesis size).

 

I'll also let you verify that

 

n₅ = (4⁵*N - sum[i=5 to 1] 4ⁱ*5^5-i )/5⁵, and that this simplifies to 

 

n₅ = (1024*N - 8404)/5⁵. We're almost done, hang in there! What this tells us is that, since n₅ must be an integer (this is what's left for the grandchildren), we have

 

1024*N - 8404 = 0 (mod 5⁵). But, we also know that n₅ must be divisible by another factor of 5, since the 5 grandchildren split n₅ equally between themselves. Thus, all told, we have

 

1024*N - 8404 = 0 (mod 5⁶=15625), or adding 8404 to both sides,

1024*N           = 8404 (mod 5⁶). Now 1024 = 2¹⁰ has an inverse mod 5⁶ (since 2¹⁰ and 5⁶ are coprime), so we multiply by that (10849) to see

 

N                   = 8404*10849 = 3121 (mod 5⁶).

 

This is our answer; he must have bequeathed 3121 gold coins total.

 

I really expected to use the Chinese Remainder Theorem, but this turned out to be the only way I could figure out how to solve the problem. 

 

Good luck!

 

 

Answer 3

We can put this into an equation. It looks messy, but it really isn't.

 

We have some number N, which is how many coins the guy had in his collection.

N

 

First, one coin is given to the butler.

N - 1

 

The eldest child gets 20% of the coins, leaving 80% in the pile.

(N - 1) x 0.8

 

The butler gets a coin.

((N - 1) x 0.8) - 1

 

The second child takes 20%

(((N - 1) x 0.8) - 1) x 0.8

 

Butler

((((N - 1) x 0.8) - 1) x 0.8) - 1

 

Third kid.

(((((N - 1) x 0.8) - 1) x 0.8) - 1) x 0.8

 

Butler

((((((N - 1) x 0.8) - 1) x 0.8) - 1) x 0.8) - 1

 

Fourth kid

(((((((N - 1) x 0.8) - 1) x 0.8) - 1) x 0.8) - 1) x 0.8

 

Butler

((((((((N - 1) x 0.8) - 1) x 0.8) - 1) x 0.8) - 1) x 0.8) - 1

 

Youngest child

(((((((((N - 1) x 0.8) - 1) x 0.8) - 1) x 0.8) - 1) x 0.8) - 1) x 0.8

 

Now the rest of the coins are divided among the five grandchildren

((((((((((N - 1) x 0.8) - 1) x 0.8) - 1) x 0.8) - 1) x 0.8) - 1) x 0.8) / 5 = X, where X is the number of coins each grandchild gets.

 

We can move all of these operations to the other side of that equality, and it starts to look an awful lot like a function.

ƒ(x) = 1.25(1.25(1.25(1.25(6.25X + 1) + 1) + 1) + 1) + 1

 

I don't give answers when I explain things, so you get to figure out the rest on your own. I would suggest using a graphing calculator. Best of luck.

Answer 4

Hi Nicholas;

x=quantity of coins.

butler...1 coin

1st child...(1/5)(x-1)...

[(1/5)x]-(1/5)

butler...1 coin

2nd child..{(1/5)[(1/5)x]-(1/5)-1}...

[(1/25)x]-(1/25)-(1/5)

butler...1 coin

3rd child...(1/5){[(1/25)x]-(1/25)}-1...

(1/125)x-(1/125)-(1/5)

butler...1 coin

4th child...(1/5){[(1/125)x-(1/125)-(1/5)]-1}...

(1/625)x-(1/625)-(1/25)-(1/5)

butler...1 coin

5th child...{(1/5)[(1/625)x-(1/625)-(1/25)-(1/5)]-1}...

[(1/3125)x]-(1/3125)-(1/125)-(1/25)-(1/5)

 

Working with the figure of 3125, I do not believe this is the initial amount of coins.  We have to consider the fact of...

(x-1)

(x-2)

(x-3)

(x-4)

(x-5)

If you multiply (5)(4)(3)(2)(1) the result is 120.

(3125)(1.2)=3750

3750+1=3751

I THINK THAT IS THE ANSWER!