Removed #5 Posted August 26, 2012 Report Share Posted August 26, 2012 I got it pretty quick, but, man, that's a mean one to ask in an interview!!! I've been asked to write code during a programming interview; I've also been handed code that was supposed to be good that was really bad. I think the idea in both the math case and getting handed bad code is the same - how well, how quickly do you think on your feet and how well, how badly do you present your response. Quote Link to comment Share on other sites More sharing options...
Mollys_mum Posted August 26, 2012 Report Share Posted August 26, 2012 You do understand those? Join the club! Kindred spirit or nerd? I doubt it, G. A kindred spirit of your variety would not have seen the 'creative' aspects to the problem - only the logic. Quote Link to comment Share on other sites More sharing options...
NicTurtle Posted August 26, 2012 Report Share Posted August 26, 2012 Ok, since we're being nerds. i was right in my explanation except that I thought there were 8 balls not 7. Quote Link to comment Share on other sites More sharing options...
Mollys Dad Posted August 26, 2012 Report Share Posted August 26, 2012 Also there is more than just one Fermat's theorem lol... Indeed! But then most people wouldn't know that or even realise that they most likely covered the n=2 case (the one that does have solutions) in school Quote Link to comment Share on other sites More sharing options...
Biggles Posted August 26, 2012 Report Share Posted August 26, 2012 G your making my brain hurt and it's a Sunday evening when we should all be relaxing and chilling out. My question is: Did Sparks get the job???? Quote Link to comment Share on other sites More sharing options...
Mollys Dad Posted August 26, 2012 Report Share Posted August 26, 2012 But I am relaxing and chilling! Quote Link to comment Share on other sites More sharing options...
persephona Posted August 26, 2012 Report Share Posted August 26, 2012 2, If one side is lighter than the other, discount the 3 heavier ones plus the odd one, then put one ball in each side. If they're level the one that's left out is broken, if not it'll be apparent which one's lighter - place the lighter one & the 'odd' one from the beginning in each side to show that one's lighter & again they've all been weighed at least once same answer but with a bit more added. It meets the criteria just didn't need saying Sorry but I am still not getting it. Maybe I am being a bit dumb, but how's that meeting the criteria? If you discard the three heavier ones + the odd one out, then proceed to comparing two of the three lighter ones, then the odd one out has obviously not been weighed. If you do the same thing and then place the lighter ball on one side and the odd one out on one side, then you have used three attempts. The way I see it, if you have to have EVERY ball on the scales at least once, it is impossible to do in two attempts. Maybe I am just not seeing the logic... Quote Link to comment Share on other sites More sharing options...
Removed #5 Posted August 26, 2012 Report Share Posted August 26, 2012 Contrary to the original directions: The balls are NOT all exactly the same, one assumes that the statement should have read that they all appear to be the same with the one exception. And again, it must either take three attempts if every ball must be put on the scale; or, if both balls on the scale at the second attempt weigh the same one assumes that the odd ball is the lighter one. In any case, while the challenge could have been phrased more accurately, the intent was obvious ... I assume ... Quote Link to comment Share on other sites More sharing options...
MollynDiesel Posted August 26, 2012 Author Report Share Posted August 26, 2012 Can't ujust look for the broken one? Lol Sent from my ST18i using Forum Runner That would've been my answer lol Quote Link to comment Share on other sites More sharing options...
MollynDiesel Posted August 26, 2012 Author Report Share Posted August 26, 2012 Well last time i try anything like this! Quote Link to comment Share on other sites More sharing options...
Removed #5 Posted August 26, 2012 Report Share Posted August 26, 2012 Oh, c'mon, pick your sense of humour up, dust it off and stick it back on. The mathocrats are gonna nit-pick unless it's phrased absolutely correctly and the rest of us are just gonna have some fun with it. An expression I think of now and again "You can't fly with the eagles if you hang with the turkeys!" Don't let the turkeys get you down! (( me included!! )) Quote Link to comment Share on other sites More sharing options...
WelshSquirrel Posted August 26, 2012 Report Share Posted August 26, 2012 I know this one! Take 3 balls and place on one side of the scales, and 3 on the other, leaving 1ball beside the scales. If the scales balance then the remainder is the lighter one. If they don't then you know its in the lighter 3 on the scales. Out of the lighter 3, weigh two of them on scales. Its the same principle as before, thus leaving you with the lightest ball. Voila! Sent from my HTC Sensation XL with Beats Audio X315e using Tapatalk 2 Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.