I think I found another answer to skewer 1, the one that was 1 — 10 — 9 — 17 — 8, but the app told me it is wrong. Could someone double-check it for me? I got 1 — 100 — 99 — 26 — 8. Is it OK? Are there any others?

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Oops! I see what I did wrong. I used 99 instead of 18 when summing to 99's left. I hate that! (I do it all the time.) Still wondering if there are any other solutions, though.

Hey Peter, I'm glad you found the error!

I believe you've found the only solution.

There are only so many possibilities to satisfy the outside boxes. The 8 can only be connected to something with a digit sum of 8.

The set of all numbers with a digit sum = 8 is shown below the 17. You'll see, similarly to when you tested the 100, that the numbers only get bigger. However, the center cannot get bigger because it equals the digit sums of its connected circles, whos digit sums are 1 and 8. The center must be 9, which eliminate all options but the 10 and 17.

Here, sd(s) means sum of digits of s.

This explanation gave me a formula as the following:

For problems of the form a->p->q->r->b with known variables a, b and unknown variables p, q, r; if there is a solution it is p=sd(a)+sd(a+b), q=a+b, r=sd(b)+sd(a+b).

Here is the proof:

a=sd(p) (1), p=sd(a)+sd(q) (2), q=sd(p)+sd(r) (3), r=sd(q)+sd(b) (4), b=sd(r) (5)

Since (1), (5), and (3), q=a+b. Now from (2) and (4) if you replace q with a+b, you get the formula above.

This means that there can be up to 1 answer to problems of the form a->p->q->r->b with a, b known variables and p, q, r unknown variables.

- aj

Thank you for sharing this proof!