Not Your Average Bag Of Marbles

We can let $a$ , $b$ , and $c$ be the numbers of balls of each of the three colors. So we know that $a+b+c=100$ . The probability of drawing one marble of each color if you draw three marbles at once is $\frac{\binom{a}{1}\binom{b}{1}\binom{c}{1}}{\binom{a+b+c}{3}}=\frac{abc}{\binom{100}{3}},$ which we know is equal to 20%, or $\frac{1}{5}$ . From this we can deduce that $abc=\frac{1}{5}\binom{100}{3}=32340.$ These two constraints, along with the fact that $a$ , $b$ , and $c$ are positive integers, are sufficient to solve the problem.

We want to find a way to split 32340 into three factors that add to 100. We start by finding the prime factorization of 32340: $32340=2^2 \times 3 \times 5 \times 7^2 \times 11$ Before we go any further, let's establish a handy lemma, the proof of which is left to the reader: $\text{The system } x+y=S; xy=P, \text{ where }S,P\in\mathbb{R}\text{, has a real-valued solution if, and only if, }S^2 \geq 4P.$ The two 7's are either part of the same factor, or of different factors. Suppose they're together. That makes their factor either 49 or 98. But 98 is clearly too large, so it would have to be 49. That would mean that the other two factors sum to $100-49=51$ , and multiply to $\frac{32340}{49}=660$ . But $51^2<4 \times 660$ , so according to our Lemma, this cannot happen. Therefore the two 7's must be in different factors .

If the 11 were together with one of the 7's, that would make one of the factors 77. Then the other two factors would have to sum to $100-77=23$ and multiply to $\frac{32340}{77}=420$ . But once again, since $23^2<4 \times 420$ , our Lemma tells us that is not possible. So the 11 is not with either 7 .

Now we can write the three factors, without loss of generality, as $a=7x$ , $b=7y$ , and $c=11z$ for some positive integers $x$ , $y$ , and $z$ . This tells us that $7x+7y+11z=100$ , or $7(x+y)+11z=100.$ Solving this linear Diophantine equation gives $x+y=8-11n$ and $z=4+7n$ for any integer $n$ , but the only value of $n$ that yields positive values for both $x+y$ and $z$ is $n=0$ . Therefore $x+y=8$ and $z=4$ . Since $x$ , $y$ , and $z$ must be made from factors of $2^2$ , $3$ , and $5$ , this gives (without loss of generality) $z=4$ , $x=3$ , and $y=5$ . And therefore, $a=21\text{, }b=35\text{, and }c=44.$

Since $a<b<c$ , we have $ab+c=779.$

I also got the same equations as Matt Enlow above: $\begin{cases} a + b + c = 100 \\ abc = 32340 \end{cases}$ And I also gathered that exactly one of $a,b,c$ is divisible by 11, because 32340 is divisible by 11, but not divisible by $11^2$ . $$ Let $0<q<100$ denote the value of one of the solutions of $a,b,c$ , then the other 2 roots of $a,b,c$ satisfy $\begin{cases} A+B = 100-q \\ AB = 32340/q \end{cases}$ Vieta's formula tells us that there exists distinct integer roots (in $X$ ) for $X^2 - (100-q)X + \dfrac{32340}q = 0$ and its (quadratic) discriminant is strictly positive, $(100-q)^2 - 4 \cdot \dfrac{32340}q > 0 \quad \Rightarrow\quad q(100-q)^2 > 129360 \quad \Rightarrow\quad 21 \leqslant q \leqslant 48$

Earlier, we have established that exactly one of the values of $a,b,c$ is divisible by 11. So we just need to test for the case where $q = 22,33,44$ .

Case 1: $q = 22$ , then 2 of the positive integer roots of $\begin{cases} a + b + c = 100 \\ abc = 32340 \end{cases}$ satisfy $X^2 - (100-22)X + \dfrac{32340}{22} = 0$ . But this yields no solution.

Case 2: $q = 33$ , then 2 of the positive integer roots of $\begin{cases} a + b + c = 100 \\ abc = 32340 \end{cases}$ satisfy $X^2 - (100-33)X + \dfrac{32340}{33} = 0$ . But this yields no solution either.

Case 3: $q = 44$ , then 2 of the positive integer roots of $\begin{cases} a + b + c = 100 \\ abc = 32340 \end{cases}$ satisfy $X^2 - (100-44)X + \dfrac{32340}{44} = 0$ . Solving this gives the other two roots of 21 and 35.

Hence, $(a,b,c) = (21,35,44) \Rightarrow ab + c = \boxed{779} .$

P(getting 1 colour each)
= 3! x (a/100) x (b/99) x (c/98)
= 6abc / (100 x 99 x 98)
= 1/5
abc = 2² x 3 x 5 x 7² x 11
Playing the combinations around a + b + c = 100 from the factors listed in abc = 2² x 3 x 5 x 7² x 11 , we get
a = 21
b = 35
c = 44
for an answer of 21 x 35 + 44 = 779