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A student from Mearns Castle made the useful connection with another problem published this month:
Aswaath from the Garden International School in Kuala Lumpur, Malaysia mentioned that the method for solving this problem connected with the method used by Gauss when he was still a young student.
Joe from Hove Park Lower School also noticed connections with other work:
If, for example, 10 mathematicians met, the first will make 9 handshakes, the second makes 8, the third makes 7 and so on until the tenth finds he has already made handshakes with everyone and so makes no more.
This gives 9+8+7+6+5+4+3+2+1+0 handshakes and this is 45.
But look at the sequence... it is the 9th triangular
(See Picturing Triangle Numbers and/or Clever Carl)
The formula for the Tth triangular number is T(T+1)/2
With the handshake problem, if there are n people, then the number of handshakes is equivalent to the (n-1)th triangular number.
Subsituting T = n-1 in the formula for triangular numbers, we can deduce a formula for the number of handshakes between n people:
Number of handshakes = (n-1)(n)/2
Jayme from the Garden International School agreed and used this insight to correct Sam's reasoning:
Joseph from Bradon Forest School and Tabitha from The Norwood School used similar reasoning:
We received many more correct solutions, including very clear ones from Siddhartha and Tasuku from the Garden International School in Kuala Lumpur, Ben from Bedminster Down School, Abhinav from Bangkok Patana School and Luke from Maidstone Grammar School. Well done and thank you to you all.
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...
Sam displays cans in 3 triangular stacks. With the same number he could make one large triangular stack or stack them all in a square based pyramid. How many cans are there how were they arranged?