Tough math / logic Question -- which is a variation of another riddle which I really like.?
404162
IF we know the coin is heavy and we have 5 scale usage , we can find out the heavy coin amist 243 coins.\n\nFor the first question, with 3 scale usage and when u dont know is the coin is heavier or liter\nNumber the coins A-L. \n\nBegin by weighing four against four. If they balance, then weigh any of the remaining three against any three of the good coins. \n\nIf they balance then the odd one is the remaining coin and can be \nidentified (whether it's lighter or heavier) in the final weighing. \n\nIf three against three do not balance, take the three containing the odd \ncoin and weigh any one against another. \n\nThus: \n\nIf the first weighing of four against four does not produce a balance, then \nthe second weighing involves three against three with the coins switched \nbetween the two pans and a good coin introduced. \n\nSo: \n\nIf A+B+C+D > E+F+G+H \n\nWe try A+B+E against C+F+J \n\nIf A+B+E + C+F+J then either D is heavier or G or H is lighter, so weigh \nG against H. \n\nIf A+B+E > C+F+J then either F is lighter or A or B is heavier, so we \nweigh A against B. \n\nIf A+B+E < C+F+J then either E is lighter or C is heavier, so weigh \neither against a good coin, eg. K against E. \n\nAn alternative second weighing is A+B+E against C+D+F, which follows \nsimilar lines to the above.
609300
This is more of a word problem or a riddle than a math problem.\n\nThe answer is 7 (SEVEN).  Remove its first letter and it becomes EVEN!
684962
a)\nthe two actions are independent, so it doesnt make any difference, so\n\nthe answer is 3/10\n\nb) (3/10)x(2/9) + (7/10)x(3/9)\n\n(if it was red on the first draw), + (if it was white on the first draw)
377482
If it's the riddle, then the answer is "language." \n\nThere are only three words in "the English language."\n\nThe third word is "language"!
781223
A then B.  \nAnd what the heck kind of question is this?
531205
2 + 2 = 4\n\nthats all I got
1102373
i wont answer.... log onto www.mathguru.com and discover solutions with explanations.
395645
Lolzz\n\nthis question is in maths dept ....\n\n if somebody answers a question with a question, who is going to answer the second question? \n'Now answer this ...i'll see what happens...
892068
Well, you got me stumped!\n\n-------------------------------------\n\nI started working it out and got pretty far.\n\nThe above given information might be incorrect.  Maybe?  I don't know.\n\nI tried it several times.  I even double and triple checked my work!\n\n---------------------------------------------\n\nThis is as far as I got and then I got stumped:\n(D-E) (B+C) = (10J + AFGH) / A\n\n------------------------------------------------------\n\nGreat question!\n\nI look forward to the continuing discussion.\n\nI love the way you worded your question.  Right on!!!\n\nKeep those challenge questions coming!  ;o}
1102516
by any chance is this a question from your maths homework? lol :D\n\nI cant work it out myself because to put it simple, im crap at maths.
922486
Is this a trick question?\n\nLOL.
295765
Yeah, Mary. I heard this one in 5th grade!\n\nNunu is a good answer too, but I think this is the simple version of the riddle.
809813
Seen this riddle at least 2 times already... getting old.... \n\nanswer is the other coin is the nickle and the first one is a quarter...
1082362
Sometimes - rarely - there are.  I have very occasionally come to the end of the puzzle with two numbers that could go anywhere in two squares, ie, 2 and 6 might be interchangeable.  Then I think I'm wrong, and will do it all again to be faced with the same outcome!
501645
I too failed math but to me this is a logic question:\n0+0=0\nWhat else cloud it be?
225658
You may have miss-typed the riddle!\nThe question should be:\n8 _ 4 9 1 _ _ 10 3 2 0\nIf that is the case then the answer is:\n8 5 4 9 1 7 6 10 3 2 0\n\nReason: This list of numbers is in alphabetical order
1146358
I’m not seeing how this is solvable.  I’m getting two possible answers.\n\nBut here’s how you should work the problem.  \n\nFirst of all, I think you pretty much have to assume that the x,y < 10.\n\nSo we know that Bill can’t tell what his two numbers are just by looking at the product.  As it turns out, there are very few products that allow this to happen: they are 12,16,18,24,36.  So you factor out all the number pairs that would make those products, and the sums are 7,8 (twice), 9, 10(twice), 11(twice), 12, and 13.\n\nSo we know that Albert is seeing one of those sums.  If he sees 7, then he knows that the only way for his numbers to add up to 7, and multiply up to either 12, 16, 18, 24, or 36 is for the numbers to be 3 and 4.  But he says nothing.  Therefore, he isn’t seeing 7.  Neither is he seeing 9, 12 or 13, for the same reason.  So, now we know the sum must be either 8,10, or 11.  And now Bill knows that.\n\nSo, if Bill were seeing 12 as the product, he would know that the numbers were 2 and 6, since they add up to 8.  But he says nothing, so he’s not seeing 12.  Likewise, if he were seeing 18, he would know that the numbers have to be 2 and 9.  But he doesn’t, so he’s not seeing 18, and he can’t be seeing 36 either.  So that means he must be seeing either 16 or 24.\n\nAnd now Albert knows this, so either Albert is seeing a sum of 8, and knows that the numbers are 4 and 4, or he’s seeing 11, and the numbers are 3 and 8.  I just don’t know how to eliminate either of those options.