Prove that 2 students live exactly five houses apart if

There are 50 houses along one side of a street. A survey shows that 26 of these houses have students living in them. Prove that there are two students who live EXACTLY five houses apart on the street.

How do I use the pigeonhole principle for this question?

Number the houses sequentially from 1 to 50. Define 5 pigeonholes using the house numbers (1, 6, 11, ..., 46), (2, 7, 12, ..., 47), ..., (5, 10, 15, ..., 50).

Since you are distributing 26 pigeons into these 5 pigeonholes, one of them receives at least 6 pigeons. Since there are 6 pigeons (i.e. 6 numbers are being chosen), it must be that two of them are adjacent. (You can make this last statement more precise with ANOTHER pigeonhole argument.)

You can also divide the houses in 25 groups of 2 houses:

$(1,6), (2,7), (3,8), (4,9), (5,10)$ and add 10, 20 , 30, 40 for the other $4*5=20$ groups.

26 students, 25 pairs....

Hint: divide the houses into groups $(1,6,11,\ldots,46), (2,7,12,\ldots,47),$ etc.

Added after the fact: Even easier: consider the pairs $(k,k+5)$ for $1 \le k \pmod {10} \le 5$. There are 25 of them. Only one application of the principle.

You can use the possible remainders after division as your holes. If we number the houses from $1$ to $50$, we can see that $10$ of them have a remainder of $0$ after dividing by $5$, $10$ have a remainder of $1$, and so on. Within each group of ten, I can pick five houses such that none of them are five units apart. This allows me to place $25$ houses ($5$ houses for each of $5$ remainders). Where does the $26^{\text{th}}$ house go?