NewScientist tantalizer 224 from 1971

I subscribed to New Scientist from age $\sim 14$ when I was at school and we had a special arrangement to get it cheaper than normal, until a couple of years ago when I finally cancelled because of the cost and the too often mediocre content. Through most of that time I did the weekly puzzle (and even won the £15 prize on one occasion). For as long as I can remember it was the Enigma, but someone reminded me that there was a different puzzle column a long time ago which preceded the Enigma.

I was searching for the identity of the series of New Scientist puzzles that preceded the Enigmas (that were terminated to my chagrin relatively recently). To my surprise I not only discovered the series name (Tantalizer) but found that many (all?) of them were available on Google Books.

Picking one at random I looked at tantalizer #224 from NS for 1971-12-16:

As there are three excursions and seven families each with a different number of members $\ge 3$ such that every family goes on one or more excursions and each excursion is full with $22$ excursees. This is equivalent to asking us to fill in the circles in the figure below with integers $\ge 3$ such that the circles in each row all have the same value in them, and each column sums to $22$. Each column corresponds to an excursion and row to a family:

Since all the columns sum to $22$ it is plausible that the smaller numbers preferentially occur in the bottom rows. So as a first guess I put from the bottom upwards: $3,\ 4,\ 5$ and $6$ in the bottom rows, and fill in the remaining circles to get the right column totals:

Now this does not satisfy one of the constraints: the Browns, a party of 6 visited only one destination, so it must be in one of the top three rows. So I move the $6$ into the row where the $8$ was and recalculate the cell that had a $9$ before to give:

Now a little work tells us that the Robinson's have 8 children (they are the party of $10$) and visited only Ronda.

What bothers me about this is the luck in getting so close to a workable solution with the first guess at assigning the numbers to circles. I think the expected method would be to assign the $6$ to a row with a single circle, and observe that the minimum is $\ge 3$ and maximum must be $9$ or $10$ and then working the table a bit like a suduko to arrive at a solution. I must admit to laziness, if I had to go down this route I would have written some code to solve the final stage.