Or search by topic
The line before 7 - 1 - 1 - 1 either had 1 or 7 heaps.
If 1 heap (of ten) then the next line would have to be 1 - 9 which isn't what we want.
But 7 heaps will work :
Three of those seven have to be twos so they'd go to ones at the next move.
That leaves four of the ten sticks for the other four spaces - so it has to be one in each place.
This means that the arrangement 2 - 2 - 2 - 1 - 1 - 1 - 1 is the only possible arrangement to go before 7 - 1 - 1 - 1Hamish from New Zealand had similar reasoning to Fiona (well done Hamish).
He then wondered about a generalisation.
If 7 - 1 - 1 - 1 can have some thing before it, will any number of heaps ( arranged as n and the rest ones ) always have something before it?
Exploring this conjecture, combinations such as 8 - 4 - 1 and 6 - 1 - 1 - 1 - 1 are possible by starting with
2 - 2 - 1 - 1 - 1 - 1 - 1 - 1 and 2 - 2 - 2 - 2 - 1 - 1 - 1 - 1 respectively.
15 = 7 + 8 and 10 = 1 + 2 + 3 + 4. Can you say which numbers can be expressed as the sum of two or more consecutive integers?
I have forgotten the number of the combination of the lock on my briefcase. I did have a method for remembering it...