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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.)
Solution One type of proof | Comment 1 of 7
Recursively:
First show for k=1
lod(2)=1

Next assume for k and show it works for k+1
Given lod(k+1, k+2, k+3,  ... 2k-1, 2k) sum to k^2
Show lod(k+2, k+3, ... 2k, 2k+1, 2k+2) sum to (k+1)^2

We are removing a term from the beginning and adding two to the end.  The sum is therefore equal to
k^2 - lod(k+1) + lod(2k+1) + lod(2k+2)
Now
2k+2=2(k+1) they have the same lod, 
and
2k+1 is odd so lod(2k+1)=2k+1
So
lod(k+2, k+3, ... 2k, 2k+1, 2k+2) = k^2 + 2k+1 = (k+1)^2

QED

I really like this problem but can't figure out any other ways of proving it.  

  Posted by Jer on 2017-09-10 10:09:21
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