In how many ways can it happen in the lottery that if the five drawn numbers are arranged in ascending order, each number is exactly the same amount larger than the previous one? - How many draws are possible in which this condition is met for any four out of the five drawn numbers?
I. Solution. a) In the case of the first question, the difference between the largest and smallest of the five drawn numbers is four times the difference $d$ of the adjacent numbers. On the other hand, the largest difference between the numbers that can be drawn is $90-1=89$, $d$ is an integer, so its largest possible value is the integer part of the quotient 89 : 4, which is 22, and every value from 1 to 22 can occur as $d$.

For a given value of $d$, denoting the first term of the sequence by $a$, the inequality $a+4 d \leq 90$ must be satisfied, that is, $a$ can take any value from 1 to $90-4 d$, and only these; their number is $90-4 d$. Since ($d$ and $a$ determine the five drawn numbers, the number of the assumed valid lottery draws for every possible value of $d$ is also $90-4 d$.

Accordingly, we need to determine the sum of the numbers $90-4 d$ for $d=1,2, \ldots 22$, that is, the sum

$$
N_{5}=86+82+\ldots+2
$$

This is an arithmetic series, and by the sum formula, $N_{5}=968$.

b) Similarly, we can determine the number of four-term arithmetic sequences that can be formed from the numbers $1,2, \ldots, 90$: the possible values of $d$ are $1,2, \ldots, 29$, for each value of $d$, the number of values that $a$ can take, and thus the number of sequences, is $90-3 d$, and their sum is

$$
87+84+\ldots+3=1305
$$

In each case, the remaining fifth number can be chosen in 86 ways, because there are always this many numbers left in the lottery ball. The resulting $1305 \cdot 86=112230$ drawing possibilities, however, are not all different. If, for example, the added number is exactly $d$ larger than the largest number in the four-term sequence, then we are faced with a drawing possibility examined in part a), and this can also be obtained by completing the sequence of the largest four terms with the number that is $d$ smaller than the smallest term. Thus, the number of drawings determined contains the drawings examined in part a) twice, so

$$
N_{4}=112230-968=111262
$$

lottery drawings are possible in which some four numbers form an arithmetic sequence, and the fifth number is arbitrary.

Solti László (Budapest, Fürst S. g. IV. o. t.)

II. Solution for part a) of the problem. We can also determine the number of the desired draws by grouping them according to their smallest number. For a given value of $a$, based on $a+4 d \leq 90$, $d$ can take every integer value from 1 that satisfies

$$
d \leq \frac{90-a}{4}
$$

that is, the largest possible value of $d$, and thus the number of draws starting with $a$ is

$$
\left[\frac{90-a}{4}\right]
$$

where the square brackets denote the integer part of the enclosed number. For the values $a=1,2, \ldots, 86$ that can occur, the sum is

$$
\begin{gathered}
22+22+21+21+21+21+20+\ldots+2+1+1+1+1= \\
=2 \cdot 22+4(21+20+\ldots+2+1)=44+2 \cdot 22 \cdot 21=968
\end{gathered}
$$

indeed, in the sum, the numbers 21, 20, ..., 2, 1 each appear 4 times, while the remainder of the division of 90 - a number by 4 takes the values $3,2,1,0$ respectively.

$$
\text { Havas János (Budapest, Berzsenyi D. g. I. o. t.) }
$$

Remarks. 1. Some people misunderstood part b) or considered only those draws in which the four smallest or four largest numbers increase by the same amount, or excluded the draws discussed in part a). We accepted their solution if they correctly solved the problem they set for themselves, although the text of the problem does not contain any such restriction.

2. It is easy to calculate that the total number of possible lottery draws is 43949 268. According to this, a lottery draw indicating a five-term arithmetic sequence is expected on average once every about 45400 draws, while a sequence of at least four terms is expected on average less than once every 400 draws. Thus, in the ongoing Hungarian lottery, such a draw is expected once in not quite 900 years, or once in 8 years.

We note that in the Hungarian lottery, which ran 373 draws by the end of April 1964, a four-term sequence occurred once: on September 6, 1957, the numbers 39, 74, 76, 78, 80 were drawn.