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'And So on and So On' printed from https://nrich.maths.org/
There were lots of correct solutions to this problem and many
different methods were employed.
Below are three of
them.
Chris, from St Bees School in
Cumbria offered two methods:
First Method
(this method was also suggested by Andrei (from School 205,
Bucharest) and Atharv ( from Bedford Modern School), both of whom
listed all the numbers.
We observe there are 4! [24] different numbers.
We can add these on their own
2457 2475 2547 2574 2745 2754
4257 4275 4527 4572 4725 4752
5247 5274 5427 5472 5724 5742
7245 7254 7425 7452 7524 7542
Second Method
We observe there are 4! [24] different numbers.
We can observe that these are symmetrical around the mean (editor's
comment: Why?) and so will add to:
1/2[S + L]n
S = smallest number, L= largest number, n=total numbers [4! here]
Both methods give the solution as 119988
Third Method
By Langran-Goldsmith, Rebecca, Caroline and Gemma (from Northampton
High School). A similar method was used by Clement.
Each digit appears six times in each position.
So you need to add together:
6 x (2000 + 4000 + 5000 + 7000) + 6 x (200 + 400 + 500 + 700) + 6 x
(20 + 40 + 50 + 70) + 6 x (2 + 4 + 5 + 7) = 119988.