On one side of a street block, from corner to corner, the sum of the house numbers is 117. What is the house number of the fifth house from the beginning of the block?
I. solution: In the conventional numbering of houses on a street, house numbers always increase by 2, so the house numbers on a continuous segment form an arithmetic sequence. On one side, all house numbers are either odd or even. Let the number of the first house on the segment be $a_{1}$, and the number of houses be $n$, then the number of the last house is $a_{n}=a_{1}+2(n-1)$, and thus their sum is:

$$
S_{n}=\frac{\left(a_{1}+a_{n}\right) n}{2}=\frac{\left[2 a_{1}+2(n-1)\right] n}{2}=\left(a_{1}+n-1\right) n=117
$$

and from this

$$
a_{1}=\frac{117}{n}-n+1
$$

Accordingly, $n$ can only be one of the divisors of 117, and specifically 5 or a larger number, since there are at least 5 houses on the street segment, and also such that $a_{1}$ is positive. Already at this point, writing 117 as a product of prime powers $117=3^{2} \cdot 13$, the divisors are: $1,3,9,13,39,117$, among which only $n=9$ meets the requirements, and with this $a_{1}=5$. Thus, the fifth house bears the number $5+4 \cdot 2=13$. (This house is also the fifth from the end of the segment.)

László Magdolna (Budapest, Martos Flóra Secondary School, 3rd year)

II. solution: The sum is odd, so it can only be about the odd side of the street segment. And the number of houses on the segment is also odd, because the sum of an even number of odd addends is even. The sum of an odd number of consecutive terms of an arithmetic sequence is equal to the product of the middle term (denoted by $m$) and the number of terms $n=2 k-1$. Thus, on the one hand

$$
m n=117
$$

on the other hand, the middle ($k$-th) term of a $2 k-1=2(k-1)+1$ term sequence in our case is at least $1+2(k-1)=2 k-1$, i.e., $m \geq 2 k-1=n$. According to this, $117=m n \geq n^{2}$, and thus $n \leq 10$. However, $n \geq 5$, so only 9 is the appropriate size among the divisors of 117 for $n$. From this, the sought house number is again 13.

Gergely Márton (Pannonhalma, Bencés Secondary School, 3rd year)

III. solution: Our row of houses is on the odd side. If we add up $n$ odd numbers starting from 1, the sum is $n^{2}$. Since 117 is not a perfect square, the first house in our row cannot be 1. Thus, 117 must be writable as the difference of two square numbers: $117=b^{2}-c^{2}$, where $b$ is the number of houses from the beginning of the street to the end of our row, and $c$ is the number of houses preceding our row. Now

$$
b^{2}-c^{2}=(b-c)(b+c)=117=1 \cdot 117=3 \cdot 39=9 \cdot 13
$$

based on the above, we select the solution similarly.

Gergely Márton (Pannonhalma, Bencés Secondary School, 3rd year)

Remarks. 1. Here, to speak simply, we assumed that there is a house on every plot. Of course, we always assumed that there are no merged or divided plots on the street segment (e.g., 11-13, or $7 /$ a).

2. Some papers also provided solutions based on other numbering principles. We do not elaborate on these here.