*We'd like to thank Syed Abbas, Associate Professor and Chairperson SBS, IIT Mandi, India, for sending in this puzzle. Reportedly, a version of it was put to Ali ibn Abi Talib in the seventh century AD. Another version appears in Fibonacci's famous Liber Abaci. *

Two people on a long walk sit down for a well-deserved break. Person A has brought along 3 cakes to eat and person B has brought along 5 cakes.

As they are just about to tuck in person C arrives and asks to share their meal. A and B agree. They cut each cake into three equal pieces and each person eats one piece of each cake.

After the meal person C pays 8 coins for the cake. Person A gives 5 of these coins to person B and keeps 3 for himself. But person B complains. She demands to be given 7 of the coins with only 1 remaining for A.

Who is right and why?

Can you generalise the solution for an initial $k$ people having $n_1, n_2, ...,n_k$ cakes respectively, who are then joined by another $m$ people who pay $n_1+n_2+n_3+...+n_k$ coins each?

## Answer

I think that person A is correct as each coin should contribute to a cake

## Answer

I think person A in corect as each coin should be worth a cake.

## Math

Person A is right because each coin should equal 1 slice.

## No bro A person also eats 2…

No bro

A person also eats 2 extra slice from person b so he should give him 2 coins extra

## Maths

Person c paid 8 coins meaning for 8/3 proportion of cake equals 8 coins.

Therefore, if we count what person A contributes it comes to 1/3 portion after subtracting his share for food. It means he should get 1 coin.

Person B is final score comes at 7/3 meaning she should get 7 coins.

## B is right

C is paying 8 coins for 8/3 cakes, of which 1/3 is contributed by A and 7/3 by B. To fairly share these coins, A should get 1 and B should get 7.

## maths

so thats correcet answer