MEI Maths Item of the Month
July 2013
Fair share
The numbers 1-10 are split into two groups of 5 such that
a1 > a2 > a3 > a4 > a5
and
b1 < b2 < b3 < b4 < b5.
Show that, for any split, |a1 − b1| + |a2 − b2| + |a3 − b3| + |a4 − b4| + |a5 − b5| = 5².
e.g.
10 > 7 > 6 > 3 > 1
2<4<5<8<9
|10 − 2| + |7 − 4| + |6 − 5| + |3 − 8| + |1 − 9| = 25
Will this work for any list of numbers of size 2n?
i.e.
|a1 − b1| + |a2 − b2| + ... + |an − bn| = n²
Solution
Only one value from 1-5 will appear in any column (and consequently only one value from 610). To show this is true consider the case where two numbers from 1-5 appear in the same
column, e.g.
3
5
The remainder of the numbers from 1-5 would have to go to the right of 3 or the left of 5 but
there are only three numbers to fill four places and therefore this is not possible. Similarly if
no numbers from 1-5 appeared in a column this means that two of the numbers from 6-10
appear in a column which is also not possible by symmetry. Hence only one value from 1-5
will appear in any column.
Each value of |a − b| can be written as a – b, if a is larger, or b – a if b is larger.
e.g.
|10 − 2| + |7 − 4| + |6 − 5| + |3 − 8| + |1 − 9| = 10 − 2 + 7 – 4 + 6 − 5 + 8 − 3 + 9 – 1
As each value of |a − b| contains one value from 1-5 and one value from 6-10 this can be
rewritten as:
10 + 9 + 8 + 7 + 6 – 5 – 4 – 3 – 2 – 1 = 40 – 15
= 25
The above result can be generalised for any list of numbers of size 2n.
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v1.0 04/10/13
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