12 days of maths

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Geoff Abell
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12 days of maths

Post by Geoff Abell » Fri Feb 21, 2020 9:51 pm

Always like a bit of maths and we had it in spades this week.

So roulette wheel sum as a short cut to me would be (1+36)*(20-2) = 666
Was there a short cur for 12 days of Christmas? 1*12+2*11...+12*1? Split at 6*7 and double but it still took me 30 +2 seconds to work it out!

Anyone do this? Am I simply getting old and slow?

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Re: 12 days of maths

Post by oddiesdave » Sat Feb 22, 2020 4:18 pm

You forgot to add the zeros for the roulette wheel.

Dave Dunford
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Re: 12 days of maths

Post by Dave Dunford » Fri Mar 13, 2020 8:04 am

Apparently what you need is the 12th tetrahedal number (the sum of the triangular numbers up to the 12th).

The formula for the nth tetrahedral number is (n x (n + 1) x (n + 2))/6.

For n = 12, this gives (12 x 13 x 14)/6 = 364

Even I had known and remembered the formula I don't think I could have done the arithmetic in 30 seconds, but I suppose it's technically possible.

https://www.mathscareers.org.uk/article ... christmas/

[In this case the calculation reduces to 13 x 14 x 2 (i.e. 26 x 14 or 13 x 28), which I suppose is a bit more feasible in 30 secs.]

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