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

(fivethirtyeight)

Solution

Reading the question charitably (since “random integer” has no specific meaning), there will be an answer if there is a limit for a uniform distribution of positive integers up to some number $N$. But we can ignore that technicality, and make do with the idealization that since every second, fourth, fifth, and twenty-fifth integer are divisible by $2, 4, 5,$ and $25$, the chances of getting a random integer divisible by those numbers are $1/2$, $1/4$, $1/5$, and $1/25$.

The product $XYZ$ is divisible by $100$ if and only if it has at least two factors of $2$ and at least two factors of $5$. The chance that it has at least two factors of $2$ is $1$ minus the chance that it has exactly zero factors of $2$ (which is $(1/2)^3$ or $1/8$) and minus the chance that it has exactly one factor of $2$. It has exactly one factor of two if one of the three numbers itself has exactly one factor of $2$ and the other two are odd. A number has exactly one factor of $2$ if it is one of the half of all even numbers that is not divisible by $4$; so $1/2$ times $1/2$, or $1/4$ of all numbers have this property. So:

The calculation of the chance that $XYZ$ is divisible by $25$ is similar. In this case, the chance that a number has exactly one factor of $5$ is the chance that it has at least one (which is $1/5$) minus the chance that it has at least two (which is $1/25$). So:

These two probabilities are independent, because every twenty-fifth number divisible by $4$ is also divisible by $25$ (and vice versa). Therefore the probability that $XYZ$ is divisible by both $4$ and $25$, that is, that it is divisible by $100$, is their product, which is exactly $.1243$.