Consider a set of all 9 digit numbers composed of 9 distinct non-zero digits (e.g. 546327891).
An unordered pair chosen from the above set may sum to
987654321.
Without performing exhaustive search, prove that the number of such pairs is odd.
I'm not sure this is a demostration:
S = [set of all the pairs of numbers composed by 9 distinct non-zero digits, whose sum is 987654321].
A= (abcdefghi) ,B= (jklmnopqr) belongs to S => i+r=h+q
Then for each A,B there will be another pair C,D con C= (abcdefgih), D= (jklmnoprq). Then C,D will also belong to S.
F. es: 123456789 123456798
864197532 864197523
------------- -------------
987654321 987654321
But if A and B have all their digits in the same order except for the last to digits, [A=(abcdefghi) B =(abcdefgih)], there is not a C, D, pair because the permutation of the last two digits is again the pair A,B.
There is a case in which this happen:
493827165
+ 493827156
----------------
987654321
This case is unique. Then the number of elements of S will be odd.
Edited on June 9, 2016, 11:56 am
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Posted by armando
on 2016-06-09 10:46:56 |
An idea