How many (infinite) arithmetic sequences of natural numbers are there, in which the ratio of the sum of the first $2 n$ terms to the sum of the first $n$ terms does not depend on the choice of $n$, and furthermore, one of the terms of the sequence is the number 1971?
The ratio of the sum of the first $2n$ terms to the sum of the first $n$ terms of any arithmetic sequence - using the usual notation - can be transformed as follows:

$$
\frac{S_{2 n}}{S_{n}}=\frac{2 n\left\{2 a_{1}+(2 n-1) d\right\}}{n\left\{2 a_{1}+(n-1) d\right\}}=\frac{4 d n+2\left(2 a_{1}-d\right)}{d n+\left(2 a_{1}-d\right)}=4-\frac{2\left(2 a_{1}-d\right)}{d n+\left(2 a_{1}-d\right)}
$$

Let's examine when the ratio corresponding to $n=1$ and $n=2$ is equal:

$$
4-\frac{2\left(2 a_{1}-d\right)}{d+\left(2 a_{1}-d\right)}=4-\frac{2\left(2 a_{1}-d\right)}{2 d+\left(2 a_{1}-d\right)}
$$

By removing the fraction and rearranging, we get the condition:

$$
d\left(2 a_{1}-d\right)=0
$$

This can be satisfied in two ways:
I. $d=0$, in which case $a_{1}$ cannot be 0, because if $a_{1}=0$, then $S_{n}=S_{2 n}=0$, and we cannot speak of a ratio;

II. $2 a_{1}=d$, and similarly to I, here $a_{1} \neq 0$.

In the case of method I, every term of the sequence is the same, so according to the second requirement of the problem, it must be 1971. In this case, the sum (without using the formula) is $S_{2 n}=2 n \cdot 1971, S_{n}=n \cdot 1971$, and their ratio is 2, which matches the value of (1). There is only one such sequence.

In the case of method II, the sequence in question is: $a_{1}, 3 a_{1}, 5 a_{1}, \ldots$, i.e., the sequence of odd natural numbers multiplied by $a_{1}$, where - according to the requirement - $a_{1}$ can only be an odd natural number. This sequence will contain the desired number 1971 if and only if $a_{1}$ is a positive integer divisor of 1971, and this condition is also sufficient because every divisor of 1971 is odd.

Now, the different divisors of $1971=3^{3} \cdot 73$ are:

$$
\begin{array}{cccc}
1, & 3, & 3^{2}=9, & 3^{3}=27 \\
73, & 3 \cdot 73=219, & 9 \cdot 73=657 ; & 27 \cdot 73=1971
\end{array}
$$

The number of these divisors is 8. Taking each of these as the initial term of a sequence, and the double of the initial term as the difference, we get as many different sequences as there are, all different from each other and from the one found in I, so the number of sequences that meet the requirements is 9.

Gábor Polyák (Budapest, István I. Gymnasium IV. grade)

Remarks. 1. In accordance with the school curriculum, we only considered first-order arithmetic sequences.

2. We could have given the answer for the case of method II without listing the divisors of 1971, the number of divisors is given by the product of the numbers that are 1 greater than the exponents in the factorization $3^{3} \cdot 73$: $(3+1) \cdot(1+1)=8$.[^0]

[^0]:    ${ }^{1}$ For a theorem regarding the number of positive divisors of a number, see: László Faragó: The Elements of Number Theory, Secondary School Special Circle Booklets, 2nd edition, Tankönyvkiadó, Budapest, 1967, p. 46.