It may help the most recent visitors to this forum to read the previous 2 pages. Lots of clues, and maybe even the correct answer are given. The Maths is definately explained.
Hope that helps a little
Mark
there's 2 kinds of pieces. 88 squares. so there are 2^88 combinations, divide by the number of 30 second periods in a year and that's the answer. ?????? Is that right? it makes sense to me, but i keep getting this huge answer.
If you want a precise answer, you better use a more precise number than 365.25 for the days/year.
365.25 yields 294,209,645,050,331,839,612
The Google calculator gives 365.242199 days/year
365.242199 yields 294,215,928,906,488,990,935
Quite a difference when you're supposed to round to the nearest year.
Also, why do people doubt the validity of the answer just because it is huge?
the Lenny Riddle solved...almost
hey, how about this?
In one year, he is able to do the board set up thing 8765.76 times
The number definitely is huge...i got around 3.530612403x10^22, which is 353061240300000000000 to all you numbskulls. yup-yup.
They didn't specify if certain combinations that are techincally used twice count as being used twice. For example: two squares, two possible colors each. White White, Black Black, and Black White would be an equation involving factorials. If one had the ability to call White Black as another possibility, then the problem would involve 2^88. If you can, sumbit two answers using these two different techniques.
p.s. the answer for the factorial way of doing it is 2.1159904247595e+130
crazay
Can you describe your factorial method?
markr wrote:Can you describe your factorial method?
I assume you were referring to MY "factorial method." and I thought this because the pieces could go (using B&W for pieces)
BBBBBBBBBBB
BBBBBBBBBBW
BBBBBBBBBWW
BBBBBBBBWWW
BBBBBBBWWWW
etc, then
WWWWWWWWWWW
WWWWWWWWWWB
WWWWWWWWWBB
WWWWWWWWBBB
WWWWWWWBBBB
etc.
but like i said i assume i'm wrong. It was just a thought
Actually I was referring to mildlyexisting's method since a number was given, but since you offered something, how does each line in your post contribute numerically to the solution (i.e. what formula are you using?).
mmmm
It's 10.30 GMT on Fri 8th April, and the Lenny Conundrum page is still showing last week's question. Usually this closes late on Wednesday, and the results / new question come out on Thursday.
Does anybody know why we're running late this week??
Not that I'm addicted to these conundrums or anything -!!!!!
M
well I just checked and ur rite its stil the old 1... its 4:33 NST or my theory is neopets just 4got or nobody got the answer (which I highly doubt) and they want to leave it there.
...no they said they wont have one this week
markr wrote:Actually I was referring to mildlyexisting's method since a number was given, but since you offered something, how does each line in your post contribute numerically to the solution (i.e. what formula are you using?).
formula...uhhh...right...formula...
I dont have a "formula" yet because i am still trying to think of a way to do this, and it doesnt use factorials, i know that for a fact (thought about it alot today)
for every 2 squares there are 3 possibilities like mildly's post said, so i think if we use that knowledge we can figure this out, right?
Caesar wrote:for every 2 squares there are 3 possibilities like mildly's post said, so i think if we use that knowledge we can figure this out, right?
It's already figured out (2^88). I'm wondering what the people who suggested a "factorial method" had in mind.
For every 2 squares there are 4 possibilities: BB, BW, WB, WW. Treating BW and WB as the same is a common oversight when counting things (particularly as it pertains to computing probabilities).
2 + 2 = 4
4 + 4 = 8
142 + 468 = 621
3762 + 8271 = ?
Any ideas?
13143
Is this the latest Lenny?
curious...
I've been trying to figure this one out for & hour...
How did you get that figure, if i may ask...
2 + 2 = 4
4 + 4 = 8
142 + 468 = 621
3762 + 8271 = ?
that makes my head hurt, because there are any number of formulas that that prblem could use. It could be that you add that sums of the last two problems and minus one from that, then add that to the total.
142+468=610+8+4-1=621
or it could be that you add 11 to the total
142+468=610+11=621
or it could be that you add ALL of the previous sums minus one to the sum of the last one. (which is different from the last one in the fact that you would add the 621, or 610 not sure which. the 8 and the 4 instead of just the 610/621 and the 8)
so i have NO idea how to do this, and any thoughts would be appreciated
~Andrew~
stapel:
Good explanation. Thanks for covering for me.
markr
