# Products equaling sums

From The Riddler, June 25, 2021, a puzzle from Matt Enlow (which was originally published in Math Horizons, although I can’t find the original):

A bag contains 100 marbles, and each marble is one of three different colors. If you were to draw three marbles at random, the probability that you would get one of each color is exactly 20 percent.

How many marbles of each color are in the bag?

It doesn’t seem like there’s enough information, but there is. I recommend not trying to solve this puzzle in your head while navigating the day care parking lot, though.

Let’s say the numbers of the three colors are x, y and z. Then the number of sets of three marbles which contain one of each color is just xyz. And the number of total sets of three marbles is ${100 \choose 3}$. So we need to find positive integers x, y, z such that x + y + z = 100 and $5xyz = {100 \choose 3}.$

It’ll help to find the prime factorization of ${100 \choose 3}$, so we expand it: ${100 \choose 3} = {100 \times 99 \times 98 \over 3 \times 2 \times 1} = {2^2 \times 5^2 \times 3^2 \times 11 \times 2 \times 7^2 \over 3 \times 2} = 2^2 \times 3 \times 5^2 \times 7^2 \times 11.$

Dividing out the 5, we need $xyz = 2^2 \times 3 \times 5 \times 7^2 \times 11.$. The three big factors – the two 7s and the 11 – look like they’ll be worrisome. So let’s spread them out: let x and y both be multiples of 7, and let z be a multiple of 11. To get them to add up to 100, we need to find a multiple of 7 and a multiple of 11 that add up to 100; it’s easy to find 44 and 56. So z = 44.

That leaves $xy = 3 \times 5 \times 7^2$, so x and y must be 21 and 35.

Alternately, we can write some code to solve it, but what fun is that?

n = 100

target = (n * (n-1) * (n-2))/6
for x in range(1, n+1):
for y in range(x, int((n-x)/2)+1):
z = n-x-y;
if target == 5*x*y*z:
print(x, y, z)


which prints out

21 35 44

as expected.

But now that we have some code, is there something special about 100? Let’s iterate over the first 1000 integers:

for n in range(1, 1000):
target = (n * (n-1) * (n-2))/6
for x in range(1, n+1):
for y in range(x, int((n-x)/2)+1):
z = n-x-y;
if target == 5*x*y*z:
print(n, x, y, z)


returns

6 1 1 4
10 2 2 6
22 4 7 11
26 5 8 13
27 5 9 13
35 7 11 17
36 7 12 17
40 8 13 19
46 11 12 23
56 11 21 24
65 13 24 28
76 19 20 37
100 21 35 44
126 31 35 60
330 82 94 154
345 86 98 161
352 78 120 154
352 91 96 165
406 101 116 189
436 109 124 203
497 124 142 231
512 128 146 238
737 161 268 308


The sequence 6, 10, 22, 26, 27, 35… isn’t in the OEIS as of this writing, although some other Riddler puzzles have made it; it may be there later. These are all numbers where not just n has a lot of factors, but also n-1 and n-2 – for example 345 = 3 x 5 x 23, 344 = 2 x 2 x 2 x 43, 343 = 7 x 7 x 7. Thus ${n \choose 3}/5$ will have many factors, and many ways to be written as a product of three integers, making it more likely that one of those will have the right sum.

I had been expecting an algebraically cleaner solution, because this reminded me of another puzzle: a bag has r red and b blue balls in it. Find r, b such that the probability that two balls picked at random have different colors is exactly one-half. In this case similar code returns the solution

4 1 3
9 3 6
16 6 10
25 10 15
36 15 21
49 21 28
64 28 36
81 36 45
100 45 55

from which we can guess that the solutions are $r = (n^2-n)/2, b = (n^2+n)/2$.