Divide Cakes

namit

So there where 3 brothers who went out for a walk with a number of cakes that their mother made for them.

They all fell asleep.
The 1st one woke up and divides the cakes in 3 and there was one left over for the sheep.
He then went back to sleep.
the 2nd one woke up and thought that all the cakes where still there, so he divided them in 3 again and there was one left over for the sheep.
He then went back to sleep.
Then the 3rd one woke up and again divided them in 3 thinking that they had not been divided before. And again there was one left over for the sheep.

They all woke up and then divided the left overs again into three and there was one left over for the sheep.

How many cakes where there to start with?

No decimal or fractions only full cakes.

Try figure this out..

low

Question not an answer.

If originally there are 3x +1 cakes. Does the second brother devide 3x number of cakes into 3 or or 1x into 3? (presuming they devide them into 3 equally sized groups)

namit

3x number of cakes into 3 and one left over.
no splitting of cakes

low

well if my interpretation is right there should be many possible correct answers.

283

is the lowest i think...?

longshanks

283/3=94 with 1 left
94/3=31 with 1 left
31/3=10 with 1 left
10/3=3 with 1 left

its either that or 4 from what i can make of it

namit

k sorry i did my calculations wrong, well done you seam to have got it.

hivizman

Is this a clearer formulation of the puzzle?

The first brother divides the cakes into three equal shares. There is one cake left over, which he gives to the sheep. The first brother takes his share (one-third) and leaves the two other shares.

Then the second brother divides the remaining cakes into three equal shares. There is again one cake left over, which he gives to the sheep. The second brother takes his share and leaves the two other shares.

Then the third brother divides the remaining cakes into three equal shares. Again, one cake is left over, which he gives to the sheep. The third brother takes his share and leaves the two other shares.

Finally, the three brothers divide the remaining cakes into three equal shares, and all three take their shares. There is one cake left over, which is given to the sheep.

On this basis, if we start with n cakes, then after the first brother takes his share we are left with 2/3*(n-1) cakes. After the second brother takes his share, we are left with 2/3*(2/3*(n-1)-1) cakes, and after the third brother takes his share, we are left with 2/3*(2/3*(2/3*(n-1)-1)-1) cakes.

For this to be divisible by three with a remainder of 1, we must have:

2/3*(2/3*(2/3*(n-1)-1)-1) = 3*k+1

Simplifying and rearranging gives:

8*n = 81*k + 65 = 80*k + 64 +(k+1)

For n to be an integer, the right hand side must divide by 8, which means that (k+1) must be divisible by 8. So k = 7, 15, 23, etc (k = 7 +8*y, where y is an integer).

Solving for k=7 gives n=79.

The first brother gives one cake to the sheep and takes 26 cakes (78/3). This leaves 52 cakes.

The second brother gives one cake to the sheep and takes 17 cakes (51/3).
This leaves 34 cakes.

The third brother gives one cake to the sheep and takes 11 cakes (33/3).
This leaves 22 cakes.

Finally, each brother takes 7 cakes (21/3) and gives one to the sheep.

If "negative cakes" are possible, it is interesting to explore what happens if the three brothers start with a total of -2 cakes.

monkey tennis

Why did all three brothers divide the cakes together again at the end when they each knew they had already taken what they thought was a third?


Also, who decides to divide cakes up when they wake up in the middle of a nap?

Pherekydes

monkey tennis wrote: »

Why did all three brothers divide the cakes together again at the end when they each knew they had already taken what they thought was a third?

They started off with 79 and the last split was with 22, so obviously they were dense and greedy.

longshanks

hivizman wrote: »

Is this a clearer formulation of the puzzle?

The first brother divides the cakes into three equal shares. There is one cake left over, which he gives to the sheep. The first brother takes his share (one-third) and leaves the two other shares.

Then the second brother divides the remaining cakes into three equal shares. There is again one cake left over, which he gives to the sheep. The second brother takes his share and leaves the two other shares.

Then the third brother divides the remaining cakes into three equal shares. Again, one cake is left over, which he gives to the sheep. The third brother takes his share and leaves the two other shares.

Finally, the three brothers divide the remaining cakes into three equal shares, and all three take their shares. There is one cake left over, which is given to the sheep.

On this basis, if we start with n cakes, then after the first brother takes his share we are left with 2/3*(n-1) cakes. After the second brother takes his share, we are left with 2/3*(2/3*(n-1)-1) cakes, and after the third brother takes his share, we are left with 2/3*(2/3*(2/3*(n-1)-1)-1) cakes.

For this to be divisible by three with a remainder of 1, we must have:

2/3*(2/3*(2/3*(n-1)-1)-1) = 3*k+1

Simplifying and rearranging gives:

8*n = 81*k + 65 = 80*k + 64 +(k+1)

For n to be an integer, the right hand side must divide by 8, which means that (k+1) must be divisible by 8. So k = 7, 15, 23, etc (k = 7 +8*y, where y is an integer).

Solving for k=7 gives n=79.

The first brother gives one cake to the sheep and takes 26 cakes (78/3). This leaves 52 cakes.

The second brother gives one cake to the sheep and takes 17 cakes (51/3).
This leaves 34 cakes.

The third brother gives one cake to the sheep and takes 11 cakes (33/3).
This leaves 22 cakes.

Finally, each brother takes 7 cakes (21/3) and gives one to the sheep.

If "negative cakes" are possible, it is interesting to explore what happens if the three brothers start with a total of -2 cakes.

it doesn't say he kept a third. the wording of these puzzles often leaves a lot of room for different interpretations