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A person cuts a loaf of bread into 10 equal pieces, then 15 and 18 equal pieces. Then the loaf of bread falls apart. What is the total number of pieces? I just cant seem to reach the correct answer. I tried to fine where will the cuts of the pieces will coincide. But my approach seems to be missing something crucial. Can u help? Thanks.

Happy
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2 Answers2

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The question is not clear. Let's look at a simpler problem, to see where the difficulty lies.

Suppose we just have a two-dimensional piece of bread, and we want to cut it into ten equal pieces, then $15$ equal pieces.

We could first make nine horizontal cuts, then $14$ vertical cuts, getting $10\times15=150$ bits of bread.

Or, we could make nine horizontal cuts, and then cut each of the ten equal pieces into two pieces, one twice as big as the other, making ten biggish pieces and ten smallish pieces, $20$ pieces in all. We get $15$ equal pieces out of this by using the ten biggish pieces, and five sets of two smallish pieces each.

So, we get very different answers, depending on how we interpret the question.

With the second interpretation, in the original problem, there's a way to do it with $30$ pieces (but I haven't tried to prove that $30$ is the minimum possible):

Let the total size of the bread be $90$ units (that's the least common multiple of $10$, $15$, and $18$), and cut it into six pieces each of lengths $1,2,3,4,5$.

You get ten equal pieces, each of size nine, by taking a size five with a size four (that's six), three size threes (that's two), three size twos with three size ones (that's two more, total, ten).

You get $15$ equal pieces of size six each by taking size five with size one (six), size four with size two (six more), and two size threes (three more, total, $15$).

You get $18$ equal pieces, each of size five, by taking a piece of size $5$ (six of these), size four with size one (six more), and size three with size two (another six, total, $18$).

Gerry Myerson
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  • Another ambiguity: Per the title, If we're using $9+14+17=40$ (not necessarily parallel) straight cuts on a 2-D bread, we can get ${40 \choose 0 } + {40 \choose 1} + {40 \choose 2} = 821$ pieces. On a 3-D bread, we can get 10701 pieces! – Calvin Lin Jun 22 '23 at 05:10
  • While the second interpretation is interesting, I don't think the "cut and rearranged into X equal pieces" reflects the problem statement. – Calvin Lin Jun 22 '23 at 05:26
  • @Calvin, whether it's $821$ pieces, or $10701$, what about the requirement for equality? – Gerry Myerson Jun 22 '23 at 06:01
  • The question in the title is different from that in the body. For the first comment, I was focusing on just the number of cuts as mentioned in the title. – Calvin Lin Jun 22 '23 at 06:03
  • @Calvin, fair enough. But as it's just one loaf of bread being subjected to all these cuts, making the $34$ cuts you suggest, you're still going to have to group pieces together to get equal pieces. I gather that $34$ is the answer on the answer sheet, but I think that just reflects the laziness of the person who set the question. $30$ cuts beats $34$ cuts, any day of the week. – Gerry Myerson Jun 22 '23 at 06:11
  • Right, I'm making the 34 cuts which "naturally" gets me the equal (grouped) pieces nicely laid out in a row without having to rearrange to get the grouping. $\quad$ FWIW With your interpretation, the case of having 2 types of cuts into $n,m$ pieces is solved, and we need at least $n+m-\gcd(n.m)$ total pieces. I'm not aware if there is a general formula for having 3 types of cuts. You have demonstrated that we can do better than the natural extension of $ n+m+p - \gcd(n, m) - \gcd(m, p) - \gcd(p, n) + \gcd(n, m, p) $. – Calvin Lin Jun 22 '23 at 06:48
  • @Calvin, there's a question somewhere on this site (or possibly on MathOverflow) about how few pieces you can cut if you want to be able to divide a cake evenly among various (finite) sets of numbers of people. My recollection is that we never proved a formula for the case of three numbers. I'll try to find it. – Gerry Myerson Jun 22 '23 at 09:40
  • @Calvin, here's one of the cake-cutting questions, with links to others: https://math.stackexchange.com/questions/1383406/minimum-cake-cutting-for-a-party – Gerry Myerson Jun 22 '23 at 09:52
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On a serious note, taking the "reasonable" interpretation of the question the the bread is essentially a 1-D line, it is essentially asking: How many distinct fractions are there if we combine the following lists (IE OP's "find where will the cuts of the pieces will coincide")

  • $\frac{1}{10} , \frac{2}{10} , \ldots , \frac{9}{10}$.
  • $\frac{1}{15} , \frac{2}{15} , \ldots , \frac{14}{15}$.
  • $\frac{1}{18} , \frac{2}{18} , \ldots , \frac{17}{18}$.

Notice that there are some duplicates, like $ \frac{5}{10} = \frac{9}{18} $.

A good way to approach this is to use the Principle of Inclusion and Exclusion.

Can you give it a try? If you get stuck, let me know what you've tried.

Remember to add 1 to the total to find the number of pieces. (Why?)

Calvin Lin
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  • i think we found the answer total number of cuts=40 there are 7 pairs of coinciding cuts so we deduct 7 from 40 and add 1 to get number of pieces = 34 – Happy Jun 22 '23 at 06:14
  • @Happy Can you generalize to cutting it up into $a, b, c$ pieces? Is there a simple formula for it? – Calvin Lin Jun 22 '23 at 06:51
  • i think the approach cannot be generalized. i found a way we can do it through prime factorization-taking prime factors of the number of pieces then finding multiples of then less than the number of pieces but essentially its the same way we formerly did the question – Happy Jun 23 '23 at 06:15
  • @Happy It can be. Think about when there are just 2 cuts. What would the answer be? How far away is it from $a+b$? – Calvin Lin Jun 23 '23 at 06:45