A man had seven married sons. Before his death, he distributed his oxen among his family members. The eldest son received a certain number of oxen, and his wife received one-ninth of the remainder. The second son received one more ox than his elder brother, and his wife also received one-ninth of the remainder. The third son similarly received one more ox than the second son, and his wife also received one-ninth of the remainder, and so on for the rest of his sons and their wives. In this way, each family received the same number of oxen. How many oxen did the man have?
I. solution: Since the sons received one more and more cows in sequence, the wives had to receive one less and less in sequence so that each couple would receive the same number of cows. However, the last wife could not receive a cow, otherwise the distribution would have ended with a remainder (,divided": completed action, did not keep any for himself). Thus, the wife of the sixth son, the one before the last, received 1 cow, which was one ninth of the then remaining, so this remainder was 9, and with the removal of 1 cow, 8 cows remained for the last couple. Each of the 7 families received this much, so the father had $7 \cdot 8=56$ cows.

To verify our result, let's also calculate the share of the first son. He received $7-1=6$ fewer than the seventh son, or the seventh couple, so 2 cows. Now we can complete the detailed distribution table; the results meet all the requirements.

| family | total | son | remainder | wife | together |
| :---: | ---: | ---: | :---: | :---: | :---: |
| I. | 56 | 2 | 54 | 6 | 8 |
| II. | 48 | 3 | 45 | 5 | 8 |
| III. | 40 | 4 | 36 | 4 | 8 |
| IV. | 32 | 5 | 27 | 3 | 8 |
| V. | 24 | 6 | 18 | 2 | 8 |
| VI. | 16 | 7 | 9 | 1 | 8 |
| VII. | 8 | 8 | 0 | 0 | 8 |

Biborka Tamás (Makó, József A. g. I. o. t.)

II. solution: Let the two consecutive sons be $A$ and $B$, with $A$ being the youngest brother of $B$, and their wives $A_{1}$ and $B_{1}$ respectively; let the number of cows each couple received be $x$. $B$ received 1 more than $A$, so $B_{1}$ received 1 less than $A_{1}$. On the other hand, $B_{1}$ received 1 less than $A_{1}$, which is one ninth of the difference in the number of cows available when it was $B_{1}$'s turn, compared to when $A_{1}$'s share was determined. This difference is $x+1$, because the total number of cows claimed by $A_{1}$ and $B$ is 1 more than that of $A_{1}$ and $A$. Thus, from the two expressions of the difference between the shares of $A_{1}$ and $B$,

$$
\frac{x+1}{9}=1
$$

From this, $x=8$, and originally there were $7 \cdot 8=56$ cows.

Mezey Ferenc (Bp., II., Rákóczi g. II. o. t.)

III. solution: Let $y$ be the total number of cows and $z$ the share of the first son. His wife, on the one hand, received $\frac{y-z}{9}$ cows from the distribution in sequence, and on the other hand, 6 more than the seventh wife, or 6, from the first solution, so:

$$
\frac{y-z}{9}=6
$$

We can also express the share of the first couple in two ways, from which:

$$
z+\frac{y-z}{9}=\frac{y}{7}
$$

The solution of the system of equations (2) and (3) is $y=56, z=2$.

Horváth Erzsébet (Pápa, Petőfi S. g. I. o. t.)

Remarks. We can also get another equation next to (3) if we extend our reasoning to the share of the second couple, which is the same as the first:

$$
z+\frac{y-z}{9}=z+1+\frac{y-z-\frac{y-z}{9}-(z+1)}{9}
$$

or

$$
\frac{y+8 z}{9}=\frac{8 y+64 z+72}{81}
$$

If we set up our system of equations by sequentially working through the sentences of the problem statement to form the expressions for the shares of the consecutive individuals, and then from these the shares of the individual couples, we will get seven expressions on both sides of the equation (4) with $y$ and $z$. The fact that these are equal, as stated in the last data, already imposes six requirements on the two unknowns. How these two unknowns can meet all the requirements of the problem, we will return to this (1. this issue, problem 921).