Remainder Multiples of 1997

Given any 1000 integers, some two of them differ by, or sum to, a multiple of 1997.

Solution


|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny

Given any 1000 integers, some two of them differ by, or sum to, a multiple of 1997.

Form the 999 sets of remainders modulo 1997, {0}, {i, 1997 - i}, for i = 1,... ,998. These sets partition all remainders. Hence two of the given integers, say x and y, have remainders which lie in one of these sets. If x and y have the same remainder, then they differ by a multiple of 1997. Otherwise, their remainders sum to 1997 and so x + y is a multiple of 1997.


[an error occurred while processing this directive]

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny
[an error occurred while processing this directive]
[an error occurred while processing this directive]