Alright, picture this.
You’re hosting a little get-together and decided to get pizza from your favorite local joint. They have a deal on one massive pizza, and will slice it any way you want. You don’t know how many of the seven pals you’ve invited are going to be coming over, but you want to make sure that no matter what, everybody gets the same amount of pizza.
How many and what size slices should you ask for the pizza to be cut into so it can be divided evenly 1, 2, 3, 4, 5, 6, 7, or 8 ways?
Simplifying
So if it’s just one person, you don’t need to slice it. We can ignore the single person case for the rest of this since you can always split it fairly when it’s just you. For two people, you can just cut it into halves.
For three people, you would want each to get a third. But if you cut into thirds, then there’s no way to split it between two people. So you split one of your thirds and end up with two thirds and two sixths for four total slices. You could split it into six sixths, but our goal is to use the fewest slices possible.
Here’s a table to show how we divided up the pizza. Each cell in each column will always equal the fraction of the total pie indicated by the header, and in total, each column equals a full pizza.
| Halves | Thirds |
|---|---|
| 1/3 + 1/6 | 1/3 |
| 1/3 + 1/6 | 1/3 |
| 1/6 + 1/6 |
For four people, you could cut the two thirds you have left into a fourth and a twelfth for six total slices.
| Halves | Thirds | Fourths |
|---|---|---|
| 1/4 + 1/6 + 1/12 | 1/4 + 1/12 | 1/4 |
| 1/4 + 1/6 + 1/12 | 1/4 + 1/12 | 1/4 |
| 1/6 + 1/6 | 1/6 + 1/12 | |
| 1/6 + 1/12 |
But if you play around with it a bit more, you find that this in not the only solution to divide the pizza for up to four people. You could instead use three fourths and three twelfths and use the same number of slices.
| Halves | Thirds | Fourths |
|---|---|---|
| 1/4 + 1/4 | 1/4 + 1/12 | 1/4 |
| 1/4 + 1/12+ 1/12 + 1/12 | 1/4 + 1/12 | 1/4 |
| 1/4 + 1/12 | 1/4 | |
| 1/12 + 1/12 + 1/12 |
One thing that we can note from working with four is that dividing by two comes automatically. If one number is a factor of another, it can be ignored since making the multiple work guarantees that the factor will work. Now that we have the basics, let’s start looking at our original question.
Trying for Five
Five is where things start to get tricky. The least common multiple of 1-5 is 60, so instead of expressing things in sixtieths, we’re going to just call a sixtieth of the pizza a minislice. A half is 30 minislices, a third is 20, a fourth is 15, and so on. To cut into fifths, we need each combination of slices to equal 12. (The pizza isn’t actually being cut into each minislice, it’s just a unit of measure.)
If we follow the line of thinking used in the first up-to-four example (which in minislices is 15, 15, 10, 10, 5, 5), we would:
- Cut our two 15’s into 12’s and 3’s
- Cut a 5 into two 2’s and a 1
This would give us a table that, in minislices, looks like this:
| Halves (30) | Thirds (20) | Fourths (15) | Fifths (12) |
|---|---|---|---|
| 12 + 12 + 3 + 2 + 1 | 12 + 5 + 2 + 1 | 12 + 3 | 12 |
| 10 + 10 + 5 + 3 + 2 | 12 + 3 + 3 + 2 | 12 + 2 + 1 | 12 |
| 10 + 10 | 10 + 5 | 10 + 2 | |
| 10 + 3 + 2 | 10 + 2 | ||
| 5 + 3 + 3 + 1 |
In total, we see ten slices. This is a great solution, except it isn’t the fewest number of slices. With a different approach, we can solve it with nine:
| Halves (30) | Thirds (20) | Fourths (15) | Fifths (12) |
|---|---|---|---|
| 12 + 12 + 3 + 3 | 12 + 8 | 12 + 3 | 12 |
| 12 + 8 + 7 + 2 + 1 | 12 + 7 + 1 | 12 + 3 | 12 |
| 12 + 3 + 3 + 2 | 12 + 2 + 1 | 12 | |
| 8 + 7 | 8 + 3 + 2 | ||
| 7 + 3 + 1 |
What Happened Here?
In short, we didn’t consider the big picture. By taking our previous solution and trying to fit it for our new problem, inefficiencies are introduced.
If we start from the ground up, we can see that we can fit three full fifths (12 minislices) into each division. The solution for fourths we were working with only had two slices bigger than a fifth, so we were already at a disadvantage. Big slices help create fewer total slices.
What seems to be effective is looking at all divisions and trying to find the largest slice you can fit into all of them. (We can ignore two since if it works for four it works for two.) So the process would look like this:
What’s the largest slice I can make that fits everywhere?
12 because nothing bigger will fit when dividing by five.
How many 12’s can I make?
Three because for four, I would need to fit two together when dividing by three and they don’t fit.
What’s the next largest slice I can make?
8 because with a 12 it adds to 20 when dividing by three.
What’s the next largest slice I can make?
7 because with the 8 it adds to 15 when dividing by four.
What’s the next largest slice I can make?
3 because with a 12 it adds to 15 when dividing by four. I can make a second for the same reason.
What’s the next largest slice I can make?
2 because with the 8 and a 3 it adds to 12 when dividing by five.
What’s the next largest slice I can make?
1 because it’s all that’s left.
How did that work?
Every step of the process, you finish a slice for some number of divisions. By using the largest slice that doesn’t overfill any slice at any size, you minimize the number of wasted cuts.
This might not be the only solution, as we saw in the four case, but it is a valid way to get down to nine slices when dividing by up to five.
Applying What We Learned to Six
When dividing by six, we will need only 10 minislices for each sixth. We are able to ignore dividing by three in addition to two since both are factors of six.
What’s the largest slice I can make that fits everywhere?
10 because nothing bigger will fit when dividing by five.
How many 10’s can I make?
Four because for five, I would need to fit two together when dividing by four and they don’t fit.
What’s the next largest slice I can make?
5 because with a 10 it adds to 15 when dividing by four. I can make a second for the same reason.
What’s the next largest slice I can make?
2 because with a 10 (or two 5’s) it adds to 12 when dividing by five. I can make four in total for the same reason.
What’s the next largest slice I can make?
1 because when dividing by four, I need to add to 10 + 2 + 2 to equal 15. This needs to be done twice, so you can make two 1’s.
All in all, that was just twelve slices.
I hope that wasn’t too confusing to follow, but I do want to stress that it’s just an algorithm you work through. You might have more success visualizing it when looking at the final product. We’ll have to move away from tables now because it just gets too wide.
- Halves (30)
- 10 + 10 + 10
- 10 + 5 + 5 + 2 + 2 + 2 + 2 + 1 + 1
- Thirds (20)
- 10 + 10
- 10 + 10
- 5 + 5 + 2 + 2 + 2 + 2 + 1 + 1
- Fourths (15)
- 10 + 5
- 10 + 5
- 10 + 2 + 2 + 1
- 10 + 2 + 2 + 1
- Fifths (20)
- 10 + 2
- 10 + 2
- 10 + 2
- 10 + 2
- 5 + 5 + 1 + 1
- Sixths (10)
- 10
- 10
- 10
- 10
- 5 + 2 + 2 + 1
- 5 + 2 + 2 + 1
Up to Seven
I won’t run through the full decision process, but it’s the same that we’ve used on five and six.
Since seven and eight don’t fit neatly into sixty, we’re going to have to use a smaller base fraction. The least common multiple of 1-8 is 840, so we’re going to call 1/840 a microslice. So a half is 420 microslices, a third is 280, etc. (To recap, a minislice is 1/60 and a microslice is 1/840. A minislice is equal to 14 microslices.)
Here is what the end product looks like for dividing by up to seven in microslices:
- Halves (420)
- 120 + 120 + 120 + 48 + 8 + 4
- 120 + 90 + 78 + 48 + 20 + 20 + 20 + 20 + 2 + 2
- Thirds (280)
- 120 + 120 + 20 + 20
- 120 + 90 + 48 + 20 + 2
- 120 + 78 + 48 + 20 + 8 + 4 + 2
- Fourths (210)
- 120 + 90
- 120 + 78 + 8 + 4
- 120 + 48 + 20 + 20 + 2
- 120 + 48 + 20 + 20 + 2
- Fifths (168)
- 120 + 48
- 120 + 48
- 120 + 20 + 20 + 8
- 120 + 20 + 20 + 4 + 2 + 2
- 90 + 78
- Sixths (140)
- 120 + 20
- 120 + 20
- 120 + 20
- 120 + 20
- 90 + 48 + 2
- 78 + 48 + 8 + 4 + 2
- Sevenths (120)
- 120
- 120
- 120
- 120
- 90 + 20 + 8 + 2
- 78 + 20 + 20 + 2
- 48 + 48 + 20 + 4
This gives us a total of sixteen slices. Notice how the 120s go until four fills up, then a 90 is added to round that up, then the 78 is added to finish off a fifth, then 48s to complete other fifths, then a bunch of 20s for the sixths. Every slice has its spot.
How Many Slices to Divide Between 1-8?
At last, we can tackle our original question. Just a note, if we had followed our original faulty logic we took from four to five, we’d end up with two more slices than needed. But following our algorithm, we can answer the question once and for all:
- Halves (420)
- 105 + 105 + 105 + 105
- 105 + 63 + 63 + 35 + 35 + 22 + 22 + 15 + 15 + 14 + 13 + 6 + 6 + 3 + 1 + 1 + 1
- Thirds (280)
- 105 + 105 + 63 + 6 + 1
- 105 + 105 + 63 + 6 + 1
- 105 + 35 + 35 + 22 + 22 + 15 + 15 + 14 + 13 + 3 + 1
- Fourths (210)
- 105 + 105
- 105 + 105
- 105 + 63 + 35 + 6 + 1
- 63 + 35 + 22 + 22 + 15 + 15 + 14 + 13 + 6 + 3 + 1 + 1
- Fifths (168)
- 105 + 63
- 105 + 63
- 105 + 35 + 22 + 6
- 105 + 35 + 22 + 6
- 105 + 15 + 15 + 14 + 13 + 3 + 1 + 1 + 1
- Sixths (140)
- 105 + 35
- 105 + 35
- 105 + 22 + 13
- 105 + 22 + 6 + 6 + 1
- 105 + 15 + 15 + 3 + 1 + 1
- 63 + 63 + 14
- Sevenths (120)
- 105 + 15
- 105 + 15
- 105 + 14 + 1
- 105 + 13 + 1 + 1
- 105 + 6 + 6 + 3
- 63 + 35 + 22
- 63 + 35 + 22
- Eighths (105)
- 105
- 105
- 105
- 105
- 105
- 63 + 35 + 6 + 1
- 63 + 35 + 6 + 1
- 22 + 22 + 15 + 15 + 14 + 13 + 3 + 1
The fewest number of unequal slices needed to divide a pizza equally between 1-8 people is 21!
Now that we have our answer, on to the real challenge… getting the pizzeria to slice it like this!
