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Lots of lods (Posted on 2017-09-10) Difficulty: 3 of 5
Let's denote the largest odd divisor of a positive integer n by lod(n).
Thus lod(1)=1; lod(72)=9; lod(2k+1)=2k+1; lod(2^k)=1.

Prove the following statement:
Sum of all values of lod(k), (k>1) from k=n+1 to k=2n, inclusive, equals n2.

Example: take n=7: lod(8,9,10,11,12,13,14)= (1,9,5,11,3,13,7), sum of the values within the last pair of brackets is 49, indeed.

Source: Shown to me as a trick i.e. "the wizard" guesses the result a priori.

No Solution Yet Submitted by Ady TZIDON    
Rating: 4.6667 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Alternate approach Comment 7 of 7 |
(In reply to Alternate approach by Paul)

I agree with chun.  This is a very cool proof, and one which explains why this works.  Thanks, very much Paul.


Now that I have seen the proof, it occurs to me that it can be simplified.  There are easier ways to say that lod(x) <> lod(y) if x and y are both between (n+1) and 2n.

None of which is meant to take away from this being a great proof.

  Posted by Steve Herman on 2017-09-11 20:33:38
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