SinisterMJ
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 Feb 18 awarded Notable Question Jun 26 awarded Popular Question Jan 21 comment 3rd grade math problem My thoughts on this was that it doesn't really promote logical thinking but doing trial-and-error. I already told her that I consider that a really dumb problem. Thanks for the help, it shows to me there was no real "formula" to solve this, but to assume no carrying, and just go from there (granted that is something I didn't do when I tried to solve it) Jan 21 accepted 3rd grade math problem Jan 20 asked 3rd grade math problem Dec 25 awarded Student Oct 4 comment “The Original IQ Test” — A peg puzzle How can the genius solution even be achieved? In order to get a Peg in a one of those 8 holes, it needed to have been in one of the other 8 holes. Thus I feel like 7 pegs is the best solution you can achieve? Oct 3 comment Riddle: 1 question to know if the number is 1, 2 or 3 @ChibuezeOpata The trick is to pose the question with a new random value the girl doesn't know, thus resulting in the "I don't know". Sep 21 comment Percentage of an amount? With Method 1 you are calculating 40% of the current pandas. Since those 6000 pandas represent 60% of the original, the 2400 are 40% of the 60%, so 24% of the original amount. Sep 10 comment Blue eyes: a logic puzzle But WHY doesn't brown eye leave after 100 days as well? Each islander sees at least 99 brown eyes, KNOWING that each of those knows that there are at least 98 brown eyes in the group? Every single brown eyed person knows that there are 99 or 100 brown eyes in the group, so that is common knowledge as well. Aug 4 comment Is $83^{27} +1$ a prime number? @JeppeStigNielsen True, that slipped my mind... Aug 3 awarded Commentator Aug 3 comment Is $83^{27} +1$ a prime number? @JeppeStigNielsen That doesn't make sense. $b^n+1$ can be a prime as well for n = 1, and b being a prime-1. So your comment cannot be entirely correct. Jul 19 comment Easter Dates Different It all depends on the accuracy needed. I feel like it will never work completely, since the rotational speeds of both earth and moon change, thus making every try obsolete in the first place. Jul 19 comment Easter Dates Different Note: leap years will be broken again after 3200 years. Thats the deal, you will have to live with a certain uncertainty, which basically screws this program. But writing a computer program to do this, no chance. Jul 19 answered Easter Dates Different Jul 19 comment Easter Dates Different Basically it should be enough to show that an earth year (365.25 days) and moon revolution (27.32 days) do not line up, resulting in a shift. And thanks to leap years, the only valid answer would be something like 400, cause for everything else you break with leap years. Unfortunately, in 400 years, the 1/1 is not the same as 1/1 + 400, so it will just not happen. Every other number can be broken with leap years. Jul 19 comment Solving a simple exponential equation This is so much easier than the other solutions. Rather straight forward. Jul 18 accepted Solve equation with variables inside and outside cos Jul 18 comment Solve equation with variables inside and outside cos Er, okay, I will refrain from trying to invent something, and instead do it numerically :)