Original can = 24L

Can A = 5 L

Can B = 11 L

Can C = 13 L

1) Fill can C from the original can

orig = 11L, Can C = 13L

2) Fill can A from can C

orig = 11L, can A = 5L, can C = 8L

3) Move the 8L from can C to can B

orig = 11L, can A = 5L, can B = 8L

4) Move the 5L from can A back to the original can

orig = 16L, can B = 8L

5) Repeat steps 1, 2, and 4

You’ll end up with 8L each in cans B, C, and the original can.

24 11 13 5

0 11 13 0 = 24

0 11 8 5 = 24

8 11 0 5 = 24

8 0 11 5 = 24

8 3 13 0 = 24

8 3 8 5 = 24

8 8 8 0 = 24

Finall all three will have 8 litres

Cos if it does then it isnt possible. Cos it would require each can to have 8 ltrs of water. But the capacity of one can is 5 lts… therefore it cant hold 8ltrs.

So I dont think all cans will have equal (even distribution ) quantity of water

to divide 24 equally between 3 is 8

if one of the cans only holds 5 , then its impossible for them to be equal

or am i missing something here

all the best

Ian