Project Euler Forum

Really interesting little problem (imho 15% is a bit low) In the mean time I know exactly what the maximum deck size for s(n) = 60 is, or any other amount of shuffles for that matter, (No, you can't shuffle this deck by hand or even forklift :D ) but the decks in between are still a mistery to me. Y...

Finally!

one minor tweak made it general use ... Phew, that was hard work. [you may visit the problems forum post #91 to make fun of me ]
Now only one to go to reach level 8

Crickey!

this was a real brainf* regarding zero based indexes and bounds and predictive computing in reachable ranges.
Brain hurts big time, but now I got the test case correct (below 1 msec)

[edit] sure enough my final answer is still wrong! I hate this problem [/edit]

Hi Ken, good hint, but my Divisor Count function is ok (got the correct result for S(1000,10) by doing it the naive way, which would take days for the original Problem) There is something obvious wrong with my S() implementation which I just don't see yet .. proofed only test cases yet, maybe there ...

I keep getting 16739 for S(1000,10), pretty fast but obviously wrong. This drives me mad :evil: I'm pretty sure my algo for S() should work though. Got the wrong result for the problems answer in about 126 seconds :mrgreen: Anyone keen to give a second pair of eyes on my python code, or experienced ...

Nope, sorry. Still don't get it. Seems that I'm a bit thick today :( [edit] The only thing I see there is that the ratios on the diagonals are the same in every matrix, but how that accounts for those lesser matrices to be squared? No clue[/edit] [edit 2] Animus was so kind to link me to the corresp...

I don't get this at all.. Can someone explain to me how these matrices relate to each other? The sum of those matrices are different and I just can't make out any other relation :oops: taking the sample: https://projecteuler.net/project/images/p420_matrix.gif Maybe I'm just a bit undercaffeinated?

well, yeah, sure. Thanks
should've seen the ease of selftesting.

Now to find out where and why the easy to spot pattern from the example values breaks

Can someone confirm the 1000th number with at least 5 Prime Factors to be 9576 ?
thank you

Got it ... very nice pattern, not easy to spot

stijn263 wrote: ↑

Sun Jan 23, 2011 6:04 pm

Can you perhaps spot a pattern among those first 22 terms?

Have fun solving!

hmm ... pattern spotting ... 1,3,10,22,63,1nn,3nn
Nope, don't see any. Any hint?

Got the same results as others for 20(correct), 25(false) and 30(false)

... way too long to get even sum P(q)=1 for q<10**3.
4600 in a bit more than an hour m(

Can't think of any other good way to optimize that anymore, there are way too many partitions to check.
There has to be a simple rule/formula for excluding terms in this, but damn, do I suck at maths

Aye, thanx a lot!

I kind of left them out by purpose.. thought once again incorrectly that a partition must consist of n>1 elements. m(
Now to make this bugger efficient. Runs way too long to get P(q)=1 for q<10**6

Resuming the problems after a long break. My terms for q <100 are [2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 81, 96] my result for P(q)=1 for q<100 is 228 (instead of 233) :shock: Has anybody any clue where I'm going wrong? My primes with P(q) = 1 are 5 7 13 17 23 43 59 61

Please remove the correct answer .. see above:

In particular don't post any code fragments or results

And well, starting with this number produces a chain with more elements than 237.. your code must have a glitch.

Yeah, please send it.. I am curious If I am able to spot the looney in an unfamiliar language.. (no guarantuee)

@Darren,

try to submit the result of a*b*c as answer..
Oh, and please remove the values of a,b and c from your post as these are the correct ones.. don't spoil the fun for others..

No, the problem description states that you need to sum them up .. another keyword is 3 or 5

Which one?

AHahahaha ... hah .. errr.

I got 23384 for p <= 100 and 23137742 for p <= 1000.
No wonder I am stuck with this