Do you know a simple unsolved math problem?
927457
Is there an infinite number of twin primes? (i.e. N and N+2 are both prime).\n\nFor other problems, see:\nhttp://mathworld.wolfram.com/UnsolvedProblems.html
1076564
What do you want to solve for? x? y? or do you want an ordered pair (x,y) that works (integers maybe?)
903397
You might try asking this question under "Mathematics."
455504
I will explain a method that is the most easy on this planet, and a unique one.\n\nI will illustrate with the help of a simple example.\n\nEg. 4-2.\n\nDraw no. line as follows:\n\n<-6--5--4--3--2--1--0--1--2--3--4--5--6--7-->\n\nNumbers left to zero are negative integers\nand to the right are the positive ones.\n\nLocate 4 on this line. Now move 2 steps behind 4 as the operation to be carried out is subtraction.\n\nThis will give answer as 2. \n\nSimilarly for addition move the steps ahead.
50930
I know it.\nIts one more than the Infinity.\nIf you know the value of infinity then substitute it and you'll get the answer.
959197
I have factored it please refer to.\n\nFrom this solution can be found
222284
duh...?  what's your question, Mam?  \n\n=0(
986918
What riddle can i solve, but you cant? =P
263164
I won 10 points!\n\nThis is pretty easy. Try to stump me with harder ones.
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}
942547
Can you please check you haven't made a mistake somewhere? I have just calculated (with 99% certainty) that there are no solutions to those equations with those restrictions.\n\nEdit - OK, I assumed you made exactly one mistake. By removing each possible word and trying to solve it, I discovered that removing 'SIGH' was the only option that gave an answer.\nt 1\nh 8\na 3\nn 7\nk 6\ns 5\ng 9\ni 4\nv 2\nwhere SIGH = 26, not 25. Then THANKSGIVING = 65.\n\nThat was actually a cooler puzzle that the original, assuming you made one mistake :)
1055731
check the statement of the problem\nand 3squared=3^2=9
1020883
i am sure you can find a math tutor that will be happy to help you when you explain what the problem is.\n\nso what IS the problem, anyway?
417502
In mathematics, Goldbach's conjecture is one of the oldest unsolved problems in number theory and in all of mathematics. It states:\n\nEvery even integer greater than 2 can be written as the sum of two primes. (The same prime may be used twice.) \nFor example,\n\n  4 = 2 + 2 \n  6 = 3 + 3 \n  8 = 3 + 5 \n10 = 3 + 7 = 5 + 5 \n12 = 5 + 7 \n14 = 3 + 11 = 7 + 7 \netc. \n\nHope that helps
934652
So let's say the problem looks like this\n\n   11\nx 12\n--------\n        <-----the answer from 11*2\n       <------answer from 11*1\n\nfirst hide the 1 from the 12 and work out the problem as 11*2\nthen work on multiplying the 1 with 11 but make sure you start putting your answer down below your first answer and the answer starts in the column that is below the 1 from the 12.\n\nfinally when the two answers are found, you can add them and get your final answer. SOOO here's what it looks like\n\n      11\n   x  12\n--------------\n           22    (this is from 11x2)\n +  11       (this is from 11 x1)\n------------\n     132 = final answer!
1001310
unanswerable question :/\n\n(Eh well you can guess.. but you cant prove it)
826583
Yes, this is a difficult one.\nI'll keep trying, but don't hold your breath.\nAll I can say so far is that I could find\nonly one solution : (x, y) = (7,1).\nThere do not seem to be any more\nsolutions with x >= 8, and this is what\nneeds to be proved. So good luck with\nfinding someone who can do this.