A elementary school student brought cupcakes to their class to celebrate their birthday! However, there's a problem: There are 25 students in the classroom, and only 24 cupcakes!

The students have recently been studying fractions. One of the students quickly puts forth a solution: "Everybody can have 24/25 of a cupcake! We can cut 1/25 off of every cupcake, combine those into one cupcake, and then we'll have 25." However, nobody seems willing to accept a "cupcake" made out of 24 cupcake crumbs.

It is up to you to figure out how to divide the cupcakes in a way that is both fair, and doesn't reduce the treats to pastry dust. What strategy can you use to ensure that the smallest slice you have to make is as large as possible?

This puzzle is purely mathematical: Every student must get exactly 24/25 of a cupcake, every cupcake is exactly the same size, the students will wait patiently while you measure and cut the cakes, etc..

This puzzle was inspired by a page from Math Curse, where the main character faces a similar situation. They resolve it by simply not eating a cupcake, but here you have no such option...

This is my first post on PuzzlingSE, so please tell me if I'm doing anything wrong!

  • 1
    $\begingroup$ If I cut every cupcake into twenty-five pieces, then give each student a 1/25th from each cupcake, then every student gets a "cupcake" made out of 24 cupcake crumbs! All fair! Nobody can complain :) $\endgroup$
    – user46002
    Nov 3, 2018 at 5:01
  • 3
    $\begingroup$ What a nice first question, welcome to PuzzlingSE! $\endgroup$
    – Christoph
    Nov 3, 2018 at 12:31
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    $\begingroup$ @Hugh - but that can't be the optimal crumbiness, because your comment contains no technobabble. Note that we can write 24 as a sum of proper divisors, 24 = 12 + 8 + 3 + 1. Therefore cut 12 cupcakes in half, 8 in thirds, 3 in eighths, and the last cupcake into 1/2 + 1/3 + 1/8 + 1/24. Further divide the leftover 1/24 by 25. Each student gets 1/2 + 1/3 + 1/8 + 1/600 = 24/25, which by the Fibonacci fraction handwave theorem is optimal, QED. $\endgroup$ Nov 3, 2018 at 19:21

4 Answers 4


The smallest slice would be:

8/25 of a cupcake

The procedure I followed was:

You ultimately need to cut at least one cupcake in 3 pieces, you cannot cut them all in half. Let's cut that one in 3 slices of 8/25, 8/25, and 9/25 size. Then you can cut other cupcakes to match the sizes of those parts. For example, if you combine the 8/25 size with a 16/25 slice from another cupcake, then you'll have a 24/25 amount of cupcake for one student. That other cupcake with have a 25-16 = 9/25 slice left-over. Match that slice with part of yet another cupcake, etc. Working further like that you will be able to give everybody 24/25 slices.

This leads to:

2 cupcakes divided as: 8-8-9, 4 cupcakes divided as 16-9, and 6 cupcakes each divided as 15-10, 14-11, and 13-12.

And the students get:

4 students get slices of sizes 8+16. Slices of sizes 9+15, 10+14, and 11+13 go to 6 students each. And the remaining 3 students get slices of sizes 12+12.


Here is the answer and i believe that this is also optimal, i will show it when i am back home:


Let's call a whole cupcake

100 unit and what we are aiming is 96 dividing these 100s into pieces

so the obvious answer from the question was


so we want as big pieces as it possible, so theoretically

two of 48 unit pieces would be ideal,

let’s try it

48+52 out of 100 units

so what we gonna do with

52? let’s find a one piece match to 96, which would be 44, divide a 100 units of cakes into 44 + 56...

and so on

100 - 48 52
100 - 44 56
100 - 40 60
100 - 36 64
100 - 32 32 36
100 - 60 40
100 - 56 44
100 - 52 48

( i will talk about why i decided to divide one piece into 3 later, you mah figure out by yourself though) and

we will have 32 units of cake left after matching up pieces into 96 units.

and we do this 2 more times

as a result, we have extra three 32 pieces, lastly combine them.

so the answer becomes


sorry, it is messy but i am on the phone at a camp site, fix it when i am back

  • 1
    $\begingroup$ You posted your 8/25 solution 30 minutes before @fishinear, but I thought they explained it better, so I gave them the checkmark. Good answer though! $\endgroup$
    – Woofmao
    Nov 3, 2018 at 16:42
  • $\begingroup$ @Woofmao interesting, there is no reasoning in that answer at all, i dont understand how you like it to be honest. $\endgroup$
    – Oray
    Nov 3, 2018 at 17:40

My smallest piece is


You split the cakes as follows:

$3$ cakes as $\frac{6}{25}+\frac{6}{25}+\frac{6}{25}+\frac{7}{25}$
$13$ cakes as $\frac{7}{25}+\frac{9}{25}+\frac{9}{25}$
$8$ cakes as $\frac{8}{25}+\frac{8}{25}+\frac{9}{25}$
This gives $9$, $16$, $16$, and $34$ pieces of sizes $\frac{6}{25}$ to $\frac{9}{25}$.

Then put them together as

$9$ cakes of $\frac{6}{25}+\frac{9}{25}+\frac{9}{25}$
$16$ cakes of $\frac{7}{25}+\frac{8}{25}+\frac{9}{25}$


How about

- Cut 19 cupcakes into five fifths each. You have 95 fifths.
- Take one fifth out of each of the other 5 cupcakes. You have 100 fifths and 5 leftovers.
- Give each student 4 fifths.
- Cut the 5 leftovers each into 5 equal pieces. Give each student 1.
Each student got the same number of the same sized bits, so it must be fair.

The smallest piece is

The fifth of the leftover, which would be 4/25 (16%) of a cupcake.

  • $\begingroup$ it seems your answer is the same as mine :) I noticed your late sorry... $\endgroup$
    – Oray
    Nov 3, 2018 at 6:21
  • 3
    $\begingroup$ @Oray it's okay, we'll just divide most of our upvotes into fifths, etc etc $\endgroup$ Nov 3, 2018 at 6:25

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