There are three houses, equally spaced on a street, labeled in order A, B, and C, in which five children live. There is at least one child in each house. The children decide on a weekend to arrange a playdate, and the meeting place is the house that minimizes the total distance traveled. What is the probability that the children meet at house B?

Solution: One can track all the possible configurations of children distributed in the three houses, along with the total distance (number of segments) travelled by the children in attending the playdate at the various houses:

We observe that the minimum total distance occurs in house B in 4 out of the 6 cases (highlighted blue), giving a probability of 2/3.

This problem is based on a 1950s Putnam problem, which involved a much more general result. For a shortcut, we note that the house that minimizes the total distance is the one in which where Child #3 resides.

Three integers a, b and c satisfy both 0 ≤ a ≤ b ≤ c and

          abc + ab + ac + bc + a + b + c = 2023

What is the smallest possible value for a + b + c?

Solution: Add 1 to both sides and factor to get (a + 1)(b + 1)(c + 1) = 2024.

We pick a = 7, b = 10, and c = 22, so that a + b + c = 39.

Three integers a, b and c satisfy both 0 ≤ a ≤ b ≤ c and

          abc + ab + ac + bc + a + b + c = 2023

What is the smallest possible value for a + b + c?

Let A, B, and C be the centers of the smaller circles, let E be the point of tangency of circles B and C, and let D be the center of the large circle, as indicated in the figure below.

Then triangle ABC is an equilateral triangle with side lengths 2r, and triangle CED is a right triangle with length of CE equal to r.  Further, because triangle ABC is equilateral, angle Θ is half of angle ACB, or π/6 .  If R is the radius of the large circle, then clearly

Now,

so that

and therefore,

Hence,