Mathematic geniuses, look here :)

[kekke2000]

I'm seeking a formula for this problem. I dont even know if it's possible.

If you have a bucket of potatoes (X). You can take a potato and cut it into 4 pieces(Y). Everytime you cut a potato you get X-1 and Y+4. Now to the fun part.

I want to know how many potatoes (X) you need to have left to have twice as many pieces(Y) than potatoes.

Example: If you have 5 potatoes and 4 pieces you only need to cut one potato to get twice as much pieces (8) as you have potatoes (4). So the answer is 4 because you will have 4 potatoes left.

X=5
Y=4
will result in Z=4

X=7
Y=8
will result in Z=6

If you can just find out how many cuts needed you can take X-(insert formula here). Please help me with this wonder!

For programmers.
I can achieve a pretty good answer from a simple VB-code, but I want a formula.

111
If X * 2 > Y Then
    X = X - 1
    Y = Y + 4
    GoTo 111
End If
Z=X

 

[Bopop4]

X=4
Y=0
will result in Z=2

Am I missing something here? I don't get this one, if you cut one potato then you will have X=3, Y=4, which isn't double.
And 8 isn't double of 2.

 

[Pupik]

Wouldn't the bucket need 6 potatoes to begin with, since: 6 - two = 4

Take the two potatoes, slice them up by four (as directed), so you get 8 pieces (Y); which is twice as many as the remaining whole potatoes (4).

Or 9 potatoes, because: 9 - 3 = 6; 3 x 4 = 12,

Basically, the number of potatoes in the bucket must be any multiple of 3.

Edit: now you want code. meh.

Sounds like the equation should be: x-(y/4)=2x

 

[lbsf1]

I'm sure that there must be a possible solution however its unlikely to be one single formula and instead be a couple of simultaneous equations.

I'm giving it a go now however it doesn't seem simple atall.

Also you have the added problem that it won't work for every potato since you need very specific numbers to get a whole number answer.

 

[MSTER232]

Tough maths equation you have there. But let me sort this out:

2 potatoes - 1 piece | Cut it once to have twice as many
3 potatoes - 2 pieces | Cut it twice to have twice as many

Z = X - Y (?)

Or let's try something else:

4 potatoes - 2 peices | Cut it twice to have twice as many.

Z = X - Y could be a possible solution.

 

[dhandeh]

It really depends on how many people you are feeding. I tend to roast them after quartering them, with a bit of garlic. In a large roasting pan toss together well the potatoes, some olive oil, salt and pepper and the garlic and then roast the potatoes in a preheated 180 degree oven, tossing them occasionally. Bake for one hour or until they are tender and golden. When ready add fresh some parsley and toss the potatoes mixture well.

 

[kekke2000]

Am I missing something here? I don't get this one, if you cut one potato then you will have X=3, Y=4, which isn't double.
And 8 isn't double of 2.

I was missing something Sharp eyes you have there

 

[kekke2000]

Wouldn't the bucket need 6 potatoes to begin with, since: 6 - two = 4

Take the two potatoes, slice them up by four (as directed), so you get 8 pieces (Y); which is twice as many as the remaining whole potatoes (4).

Or 9 potatoes, because: 9 - 3 = 6; 3 x 4 = 12,

Basically, the number of potatoes in the bucket must be any multiple of 3.

Edit: now you want code. meh.

Sounds like the equation should be: x-(y/4)=2x

Maybe you have something here but what would Z be? (the amount of potatoes that will be in the bucket when X=2Y).

 

[Apok]

So if X= the amount of potatoes
Y= amount of pieces
Z= potatoes you are left with

Z= 2X/3 + Y/6

Why?
Because

Z= X - Xu Where Xu= amount of potatoes used

Xu= 2Z/4 - Y/4

2Z is the amount of pieces you need, however you only need one potato for 4 pieces which is why it's divided by four. You also need to substract the amount of pieces you already have and also divide that by four.

Now you have

Z= X - 2Z/4 + Y/4
Z= X - Z/2 + Y/4
Multiply the equation by two
2Z= 2X - Z + Y/2
carry the Z over
3Z= 2X + Y/2
Divide by three and there you have it

Z= 2X/3 + Y/6

 

[Danoff]

X = original number of potatoes in the bucket (not the number you have currently)
Z = Number of potatoes sliced (not the number of slices, and not the number of times you sliced something)

Z = X/3

Start with 6 potatoes, Z would equal 2. You need to cut 2 potatoes into 4 slices in order to have twice as many slices (8) as you have potatoes remaining (4).

If you have 8 potatoes, Z would be 2.666. You need to cut 2.666 into 4 slices per potato to have twice as many slices (10.666) as potatoes remaining (5.333).

How did I get there.

Twice the Number of potatoes remaining = Number of slices.

Number of potatoes remaining = X-Z
Number of slices = 4Z.

2(X-Z)=4Z
2X-2Z=4Z
2X=6Z
X=3Z
Z=X/3

Other parameters you might be interested in:
Number of cuts you made = 2Z
Number of potatoes you currently have = X-Z
Number of potato slices you currently have = 4Z

 

[MSTER232]

So if X= the amount of potatoes
Y= amount of pieces
Z= potatoes you are left with

Z= 2X/3 + Y/6

Z= 2X/3 + Y/6

Yeah that sounds about right. Well done 👍 .

 

[kekke2000]

So if X= the amount of potatoes
Y= amount of pieces
Z= potatoes you are left with

Z= 2X/3 + Y/6

Why?
Because

Z= X - Xu Where Xu= amount of potatoes used

Xu= 2Z/4 - Y/4

2Z is the amount of pieces you need, however you only need one potato for 4 pieces which is why it's divided by four. You also need to substract the amount of pieces you already have and also divide that by four.

Now you have

Z= X - 2Z/4 + Y/4
Z= X - Z/2 + Y/4
Multiply the equation by two
2Z= 2X - Z + Y/2
carry the Z over
3Z= 2X + Y/2
Divide by three and there you have it

Z= 2X/3 + Y/6

You are the one I was looking for It seems to work!

 

[MSTER232]

You are the one I was looking for It seems to work!

Hopefully the GTPlanet community help you solved your lovely mathematical problem with slicing potatoes . Now have fun cooking with them 👍 .

 

[Danoff]

So if X= the amount of potatoes
Y= amount of pieces
Z= potatoes you are left with

Z= 2X/3 + Y/6

Why?
Because

Z= X - Xu Where Xu= amount of potatoes used

Xu= 2Z/4 - Y/4

So if started with 6 potatoes and cut two, you now have Z=4 and Y=8. So that means Xu should be 2. Your equation says Xu = 0.

Using your definitions of the variables:

X=Z+Y/4
Xu=Y/4

You want to set Y=2Z to have twice the number of slices as whole remaining. Subbing in:

X=3Z/2

or

Z=2/3X


Solving for the number of potatoes sliced (X-Z) in terms of the original number of potatoes X:
X-Z=Y/4
Y=2Z
X-Z=2Z/4

from above:
Z=2/3X

So X-Z=2(2/3X)/4

or

X-Z=X/3 (which is the equivalent of my post above)

 

[Apok]

No, you misunderstood. Y does not equal the number of slices you get from slicing the potatoes it is the amount of slices which are already given.
In the OP the first example has 5 potatoes and 4 pieces so X=5 and Y=4. You need to cut one of the potatoes to get 4 more slices which is Y+4.


In your case with X=6 Y actually equals to 0 in my equation so you'd get Z= 2X/3 and I believe this is what you meant with Z=2/3X (can't get to that using X=3Z/2, but you can get Z=2X/3)

 

[kekke2000]

If you feed this formula with data that wont result in a practical answer it still gives you a number.

For example if I have 3 potatoes and 30 slices I should keep slicing until I have 7 potatoes. I guess you could achieve this answer with well fitting slices and a glue stick

 

[Danoff]

If you feed this formula with data that wont result in a practical answer it still gives you a number.

For example if I have 3 potatoes and 30 slices I should keep slicing until I have 7 potatoes. I guess you could achieve this answer with well fitting slices and a glue stick

If you have 3 whole and 30 slices, you started with 10.5 potatoes. You should cut:

Z=X/3=3.5 potatoes to have twice as many slices (14) as potatoes (7).

If you've already cut down to 3, you're out of luck.

 

[Danoff]

No, you misunderstood. Y does not equal the number of slices you get from slicing the potatoes it is the amount of slices which are already given.

Ok.

In your case with X=6 Y actually equals to 0 in my equation so you'd get Z= 2X/3 and I believe this is what you meant with Z=2/3X

Agreed.

So then to go from my equation into yours, I'm giving mine a lower-case d and yours a lower-case a:

Translation:
Xd=Xa+Ya/4
Zd=Xa+Ya/4-Za

My equation:
Zd=Xd/3

Substitution:
(Xa+Ya/4)-Za=(Xa+Ya/4)/3
-Za=-2/3(Xa+Ya/4)
Za=2/3Xa+Ya/6

So our equations are equivalent.

 

[Villain]

Yay! Intermediate Algebra!

 

[MSTER232]

Yay! Intermediate Algebra!

I would actually say a problem like this is actually intermediate-advanced.

 

[terminator363]

Yay! Intermediate Algebra!

👍

 

[Villain]

I would actually say a problem like this is actually intermediate-advanced.

In California this is textbook Algebra II in high school, or Intermediate Algebra in college.

Required for undergrads 👍

 

[Apok]

Ok.



Agreed.

So then to go from my equation into yours, I'm giving mine a lower-case d and yours a lower-case a:

Translation:
Xd=Xa+Ya/4
Zd=Xa+Ya/4-Za

My equation:
Zd=Xd/3

Substitution:
(Xa+Ya/4)-Za=(Xa+Ya/4)/3
-Za=-2/3(Xa+Ya/4)
Za=2/3Xa+Ya/6

So our equations are equivalent.

Oh I see, I didn't see you had a different definition for Z.
Agreed.

 

[MSTER232]

In California this is textbook Algebra II in high school, or Intermediate Algebra in college.

Required for undergrads 👍

 

[Danoff]

In California this is textbook Algebra II in high school, or Intermediate Algebra in college.

Required for undergrads 👍

From Algebra II I remember plotting a lot of non-linear functions, finding roots, factoring quadratic equations, stuff like that. This is easier than that. At the end of the year, I think we even started messing around with some basic trigonometric functions.

I'd call this High School Algebra I, which is appropriate for 8th and 9th graders. I'd say this is appropriate for 13-year-olds.