# Let A = {8,9,10,...,2000} be a set and B be a subset of A which contains 998 elements, every two of them being distinct.How to show that, in B, are at least two elements which have the sum equal to...

Let A = {8,9,10,...,2000} be a set and B be a subset of A which contains 998 elements, every two of them being distinct.

How to show that, in B, are at least two elements which have the sum equal to 2008? thankyou.

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### 2 Answers

There are 1993 numbers in set A ranging from 8 to 2000. It is evident that with the sole exception of the number 1004, all other numbers in the set have a conplementary number in the same set A which will add to 2008.

Because 1004 is the only number without a complementary number for this question, there are ( 1993 - 1 ) / 2 = 996 pairs of such complementary numbers which add to 2008.

If set B has 998 numbers from set A, it must have at least one number from each of the 996 pairs, the number 1004, and then an additional number which will complete the complementary pair; thus two numbers from set B must add to 2008.

A = {8,9,10,...,2000}

If B containing 998 numbers is a subset of A the a subset B with the lowest sum of any two numbers will have the lowest value elements of set A and may be written as:

B = {8,9,10,....,1000,1001,1002,1003,1004,1005}

and we have two distinct elements 1003 and 1005 that have a sum of 2008.

whereas a subset of A with highest value elements will be:

B = {1003,1004,1005,...,2000}

and this subset also contains two distinct elements 1003, 1005 having a sum of 2008.

All other sets are made up elements having intermediary vulues and so will contain two distinct numbers having a sum of 2008.