Question 770986
The monkey who ate the least may have eaten {{{3}}} nuts.
That would leave {{{25-3=22}}} 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 {{{22}}}, and we cannot use {{{1}}} or {{{3}}}.
{{{5+17=22}}}
{{{7+15=22}}} and
{{{9+13=22}}}.
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 {{{3}}} sets
{3,5,17}, {3,7,15} and {3,9,13}.
 
If the monkey who ate the least eat more than {{{3}}}nuts, he/she may have eaten {{{5}}} nuts.
In that case the other two monkeys ate the other {{{25-5=20}}} nuts. We need to odd numbers that add up to {{{20}}} and we cannot use neither {{{1}}}, nor {{{3}}}, not {{{5}}}.
It could be
{{{7+13=20}}} or {{{9+11=20}}}.
That accounts for {{{2}}} more solutions, the sets {{{5,7,13} and {5,9,11}.
 
The monkey who ate the least nuts cannot have eaten {{{7}}} or more nuts, because the other two monkeys would have had to eat at least {{{9}}} and {{{11}}} nuts, and then the total would be at least {{{7+9+11=27}}} nuts.
 
So there are {{{3+2=highlight(5)}}} 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.