Problem 261

[hammingweight]

Is 1 a square-pivot?

0² + 1² = 1²
and 1 is a positive integer. 0, of course, isn't a positive integer but the wording of this problem seems to require that the square-pivot k be positive; not necessarily that the m consecutive integers immediately less than k all be positive.

Despite this, I assume that 1 isn't actually a square-pivot.

[rayfil]

I assume that 1 isn't actually a square-pivot.

That assumption is correct.

[TripleM]

0² + 1² = 1²
and 1 is a positive integer. 0, of course, isn't a positive integer but the wording of this problem seems to require that the square-pivot k be positive; not necessarily that the m consecutive integers immediately less than k all be positive

You are correct; the problem does not require the integers less than k be positive. However, the example you gave is invalid - it has n=0, but you require n >= k.

[naadv]

Hi,
Oh, Tough problem. One of the hardest I've tried.
Can somone please confirm that the solution for 10^7 is 874336206?
Thanks in advance.

[dnosrc]

Hi,
Oh, Tough problem. One of the hardest I've tried.
Can somone please confirm that the solution for 10^7 is 874336206?
Thanks in advance.

sorry but you are not even close

[Thierry]

I'm currently trying to tackle this interesting riddle, but I experience some troubles. I've read the forum so that to check I'm on the good track and verified I'm right for 10⁶.

But my solution for 10¹⁰ is still wrong, so I miss few singularities for some large values of m!

For 10⁸, I've found ???xxxxxxx?? and the sum of digits is ??
For 10⁹, I've found ???xxxxxxxx?? and the sum of digits is ??

Are these correct?

For 10¹⁰, I've found a number x of pivots, and the sum of digits of x is ??. Any clue to progress because I'm now stuck!

[sjhillier]

Thierry, your post has been edited, and I've sent you a PM.