A Father’s Fair Division

A man owned $36$ houses, numbered $1$ to $36$. The rent from each house, per month, was £100 times its number: house $1$ brought in £100, house $2$ brought in £200, and so on up to house $36$ at £3600. He had six sons, and wished to leave each son exactly six houses so that all six sons drew equal rent. How should he divide them? And if instead he had owned $100$ houses, numbered alike, to leave equally among ten sons, ten houses each?

Solution

The rents are all £100 times a house number, so equal rent means equal totals of house numbers. The numbers $1$ to $36$ add to $\tfrac{36 \times 37}{2} = 666$, which split six ways is $111$ each. So each son’s six houses must have numbers summing to $111$, and then each draws £11,100 a month.

The division almost writes itself once you pair the houses from the ends inward: $$1 + 36 = 2 + 35 = 3 + 34 = \dots = 18 + 19 = 37.$$ There are eighteen such pairs, each summing to $37$. Hand each son three whole pairs: that is six houses summing to $3 \times 37 = 111$, exactly the equal share. Any way of dealing the eighteen pairs three-to-a-son works.

The hundred-house version is the same idea. The numbers $1$ to $100$ add to $5050$, so each of the ten sons needs houses summing to $505$. Pairing from the ends, $1 + 100 = 2 + 99 = \dots = 50 + 51 = 101$, gives fifty pairs of $101$; five pairs to a son is ten houses summing to $5 \times 101 = 505$. The trick works whenever the houses pair into equal sums and the pairs deal out evenly among the heirs.