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
Among them there are 16 pairs can from 104 are:
49, 55 (=3*16+1, 103-3*16)
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.
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