Another math problem

[Jos Elkink]

Ok, similar to Thomas, I also want to put one up ... This one I stumbled on recently, and although I can manage to find the answer, I cannot manage to prove it. So if anyone can, I'd be really interested ...

Two brother sell a bunch of sheep. The amount of money they receive per sheep is equal to the number of sheep they sold. They want to divide the money between them and work as follows. First, the oldest takes 10 coins. Then the younger one takes 10. Then the older one, etc. In the end, the younger one takes what is left, after which the oldest gives the youngest one his knife, to make sure they each had an equal income out of the deal.

Question: what was the knife worth?

[thingnumber2]

ummm....more than 1?

[Oasis]

Not if there are 10 sheep - it would be worth 0.
I get a different number for every different value, almost.
For three values, the knife is worth 4, for three different values, the knife is worth 9, but sometimes 5, 0, 6, 1.
Maybe I'm just too tired, and I'm doing it all wrong?
I'll have another look tommorrow, if I can get rid of the headache this has caused.
Or, the knife is always worth 3.6 days worth of resources.

[watermelonnose]

The value of the knife is 2 coins.

The two brothers have equal income up until the last round so we can ignore the previous rounds. We know that since the older brother takes the first 10 coins and has to take the last full set of 10 coins the number of full sets of 10 coins is odd. We also know that there are at least (30+a) coins because the younger brother takes at least one set of 10 coins.

x=value of knife
a=number of coins younger brother takes on last taking

last transaction
older brother 10-x=a+x younger brother
2x=10-a a needs to be even for a whole coin value for the knife.

so we have a=2. 4. 6 or 8

We don't know what the brothers received for the sheep except that the amount is a square and greater than 30. Surveying squares we find that the only squares that have an odd tens digit (controls the odd number of 10 coin takings) also have a number 6 for the units digit.

6*6=36 14*14=196 16*16=256 24*24=576 26*26=676
OK! I haven't checked everyone and I don't know the number theory!

Therefore the younger brother on his last taking took 6 coins.

2x=10-6
x=2 Knife is worth 2 coins.

[Thomas Pickert]

Just to make your correct answer complete.

Proposition
Let m = n*10 + 4 a square number for an arbitrary integer n.

Then, n is an even number.

Proof
m is a square number, hence exists an integer s satisfying s = k*10 + 2, and m = s^2 for a particular integer k.
We have
100(k^2)+40k+4 = (k*10 + 2)^2 = s*s = m = n*10 + 4.

Obviously: n = 2*(5(k^2)+2k), hence n is even.

qed

[watermelonnose]

Jos, after sleeping on it I came up with the reason why the number 6 would have to be in the units place.

It was determined that the units number had to be even and the tens place had to be odd. If we look at the even numbers less than 10 we have 2, 4, 6, and 8

If we look at the factors

2= 2

4= 2*2

6= 2*3

8= 2*2*2

When we square with these numbers in the units place the only way an odd number is possible in the tens place is with the 6 since 6 has an odd factor.

6*6=(2*3)(2*3)=(3*3)(2*2)=9*4

[Meh]

If you change the problem then the solutions above could be used.

Nowhere in the problem does it state that an individual sheep is worth an exact coin amount. Ex: A dozen sheep for 35 coins.

Nowhere in the problem does it state that the knife is worth an exact coin amount. Ex:A knife worth 3.5 coins.

[Thomas Pickert]

The amount of money they receive per sheep is equal to the number of sheep they sold.

In the end, the younger one takes what is left, after which the oldest gives the youngest one his knife, to make sure they each had an equal income out of the deal.

The older one takes 10 coins n times. Both brothers have an equal income after the knife is exchanged. Let k be the knife's value. Let r the number of coins the younger one takes in the end.

That yields:
n*10 - k = (n-1)*10 + k + r

which is equivalent to
10 - r = 2k.

And here's David correct. The younger brother did take a coin, and we may also assume, that the knife indeed has a positive value.

Then, possible values for k are 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5.

That's it for now. I wonder if there's also a unique solution, when we allow non integer knife values.

Thomas

[watermelonnose]

If you change the problem then the solutions above could be used.

Nowhere in the problem does it state that an individual sheep is worth an exact coin amount. Ex: A dozen sheep for 35 coins.

Nowhere in the problem does it state that the knife is worth an exact coin amount. Ex:A knife worth 3.5 coins.

It is assumed since the only unit of measure is the coin. There are no half coins available. Unless you want to dull your knife by cutting one in half, but then the knife won't be worth as much.

Each sheep would be worth (2.91666666666666... coins by your example)
The problem said that the amount of coins you received per sheep is the same as the number of sheep. The number of coins per sheep is a whole number.

The knife has a unique value so we know that it needs to be a value represented by a whole number, since fractional numbers will not yield a unique answer. The problem states that the knife transfer equalized their income.

[Meh]

Thank you Thomas.

It is a bad assumption to every consider a problem in a non-analog way.

I live in a world where you can buy 2 2 liters of coke for $2.95 {not really the best deal}. Pennies are the smallest coins but even then how much is each 2 liter bottle worth.

Measure to the micrometer...
Mark with chalk...
Cut with an axe...

Life IS analog...

[rklenseth]

Hmmm.....I wonder what the worth of the coins are? And how many coins are exchanged between the brothers? And how can you know how many sheep they sold when you don't know how many coins they got?

I was always told I think to much when it comes down to math.
That is why I like history and English.

[ephiroll]

...amen for history and english...

[kroner]

Surveying squares we find that the only squares that have an odd tens digit (controls the odd number of 10 coin takings) also have a number 6 for the units digit.

6*6=36 14*14=196 16*16=256 24*24=576 26*26=676
OK! I haven't checked everyone and I don't know the number theory!

let n be number of sheep.
let n=10*k+u, where u is the units digit of n.
n^2=100*k^2+20*ku+u^2
The 100*k^2 and 20*ku terms have no effect on whether the tens digit of n^2 is even or odd because they are multiples of 20. Therefore if the tens digit of u^2 is odd, so is the tens digit of n^2. Tf the tens digit of u^2 is even, so is the tens digit of n^2.
u=0, u^2=00
u=1, u^2=01
u=2, u^2=04
u=3, u^2=09
u=4, u^2=16
u=5, u^2=25
u=6, u^2=36
u=7, u^2=49
u=8, u^2=64
u=9, u^2=81
The only values of u that will lead to an odd tens digit are 4 and 6.

Back to n^2=100*k^2+20*ku+u^2.
The 100*k^2 and 20*ku terms have no effect on the units digit of n^2 because they are multiples of 10. Therefore the units digit of n^2 is the units digit of u^2. For both u=4 and u=6, the units digit of u^2 is 6. Therefore the units digit of n^2 is 6. so the younger brother gets 6 coins at the end and the knife is worth 2 to make up for the difference.

dunno if this was done already but here is a proof of that part of watermelonnoose's answer. This also allows for non integer values of the knife by allowing for an odd number of sheep, but it shows that the only solution is still 2 (assuming the knife doesn't have zero value).

[rklenseth]

But how do you determine how many sheep there are when they don't tell you how many coins there are? It says the amount of money received is equal to the amount of sheep. But they don't tell you how many coins the brothers earned.

First, the oldest takes 10 coins. Then the younger one takes 10. Then the older one, etc. In the end, the younger one takes what is left, after which the oldest gives the youngest one his knife, to make sure they each had an equal income out of the deal.

So the oldest has 10 coins and then the youngest has 10 coins and then the oldest has 10 coins but then says ect. meaning there is more dividing up here of the coins but it doesn't tell you how many times the brothers divide afterwards.

You can't figure out the answer because the question is incomplete thus there is no answer. You need to know how many coins are exchanged between the brothers in order to figure out how much the youngest is owed and then you can figure how much the knife is worth through the exchange. Maybe, I'm just missing something. I was never good at math anyways other than logic.

[kroner]

But how do you determine how many sheep there are when they don't tell you how many coins there are? It says the amount of money received is equal to the amount of sheep. But they don't tell you how many coins the brothers earned.

I thought the same thing when I first saw the problem, but it says the number of coins they recieved per sheep is equal to the number of sheep they sold. So the number of coins they got was the number of sheep squared. Limiting the number of coins to a perfect square gives you the added info you need.