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Integer Forms (Posted on 2012-01-07) Difficulty: 2 of 5
Let us denote F(n) = Σi= 1 to n i*(-1)i+1

For example, F(4) = 1-2+3-4, F(5) = 1-2+3-4+5, F(6) = 1-2+3-4+5-6... and, so on.

(1) Determine the general form of two positive integers x and y that satisfy:
F(x) + F(y) + F(x+y) = 2012

(2) Can you explain why no positive integer solution exists whenever F(x) + F(y) + F(x+y) = 2011?

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Observations (partial solution) | Comment 1 of 2
Putting n=1 to 2000 in an excel sheet, I've noticed a pattern:

When n is odd, F(n)=(n+1)/2
When n is even, F(n)=-n/2
For question (1), if x=2011, and y=0, then F(x)=1006, F(y)=0, and F(x+y)=1006, for a total of 2012.

Furthermore, if we increase y by 2, then F(y) lowers by 1, and F(x+y) increases by 1.

Since we can do this indefinitely, and it leaves the total the same, the solution to part 1 is x=2011, and y=any even number (or vice versa).

For part (2), I can't find a solution, but I can't explain why there's no solution... Yet.

  Posted by Dustin on 2012-01-07 18:21:48
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