And so on and so on
If you wrote all the possible four digit numbers made by using each
of the digits 2, 4, 5, 7 once, what would they add up to?
Age
11 to 14
Challenge level
If you wrote all the possible four digit numbers made by using each
of the digits 2, 4, 5, 7 once, what would they add up to?
How many times can each number appear in each column when you have written all combinations of the digits down?
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.