Three numbers, 100 factors - The Snow In The Summer or So-So

Given any three random integers — X, Y and Z — what are the chances that their product is divisible by 100?

To make a number divisible by 100, there must be factors of 2, 2, 5, 5 somewhere in the three multiplicands. There can be other factors - we're partial to a tasty 37 ourselves - but any combination of numbers with two 2s and two 5s as prime factors will be divisible by 100.

Let's think about factors of 2. There's a 1/2 chance that X will have no factors of 2, 1/4 that X will have precisely one factor of 2, and 1/4 that it will have two or more factors. The same applies for Y and Z, of course.

So these arrangements have the required factors of 2:

X Y Z prob 2 0 0 1/16 0 2 0 1/16 0 0 2 1/16 1 1 0 1/32 1 0 1 1/32 0 1 1 1/32

As these are mutually exclusive possibilities, we can add them up. 9/32 combinations have the required factors of 2.

We can run a similar analysis for factors of 5: a 4/5 chance of no factors, 4/25 of precisely one factor, and 1/25 of two or more factors. Plugging these into the table gives:

X Y Z prob 2 0 0 16/625 0 2 0 16/625 0 0 2 16/625 1 1 0 16/3125 1 0 1 16/3125 0 1 1 16/3125

That's a total of 288/3125 with the required factors.

And, as the factors of 2 are independent from the factors of 5, we can just multiply these probabilities together. The final answer: 2592/100000.

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