Question 770986: Three monkeys ate a total of 25 nuts. Each monkey ate an odd number of nuts and the three odd numbers were all different and bigger than one. How many different solutions are ther?
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! The monkey who ate the least may have eaten  nuts.
That would leave  nuts for the other two monkeys.
Along with the 3 for the least hungry monkey, we need two different odd numbers that add up to  , and we cannot use  or .
 and
 .
Since I cannot tell the monkeys apart, and I would not care which one ate the least and which one ate the most, the solutions I have found so far are the sets
{3,5,17}, {3,7,15} and {3,9,13}.
If the monkey who ate the least eat more than nuts, he/she may have eaten  nuts.
In that case the other two monkeys ate the other  nuts. We need to odd numbers that add up to  and we cannot use neither , nor , not .
It could be
 or  .
That accounts for  more solutions, the sets  or more nuts, because the other two monkeys would have had to eat at least  and  nuts, and then the total would be at least  nuts.
So there are  solutions, unless you can tell the monkeys apart, and care which one ate more or less than the others.
If 3, 5, and 17 for monkeys A, B, and C respectively is considered different from 3, 5, and 17 for monkeys B, A, and C respectively, then there are 6 different ways that we can order the set {A,B,C}, and that multiplies times 6 the number of solutions.
|
|
|
