Posted by **Lolita** on Wednesday, June 22, 2005 at 9:00pm.

I need help solving this problem, I have followed the sequence forward and backwards but can't seem to find the solution. I believe it's the pigeon hole theorem. Here's the problem;

Let A be any set of twenty integers chosen from the arithmetic progression 1, 4, 7, ...,100. Prove that there must be two distinct integers in A whose sum is 104.

100 = 1 + 3(n-1) = 3n -2,

n = 34

Totally, there are 34 terms from

1,4,..,100

Among them there are 16 pairs can from 104 are:

1

4, 100

7, 97

10, 94

...

46, 58

49, 55 (=3*16+1, 103-3*16)

52

as (boy girl)

If no two chosen numbers whose sum

is 104, then we only can choose one

from the above pairs.

That is, we can at most choose

16+1+1 = 18 numbers among 34 where

no two whose sum is 104.

However, we choose 20 > 18 numbers

so at least two with sum 104.

This answer may be late, since I saw

this question a short while ago.

Kenny

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