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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.)
Some Thoughts re: One type of proof | Comment 3 of 7 |
(In reply to One type of proof by Jer)

Proving by induction, wherever applicable is a legal and reliable method.    Nobody needs another proof.

I've erroneously introduced an erroneous and redundant constraint i.e. k>1,  not being aware that the formula applies to k=1 as well.
Sorry about that.
I second SH's motion: your presentation is short and elegant.  

  Posted by Ady TZIDON on 2017-09-11 03:19:15
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