who asked the question  2+2 =5 ? tell me at priyavaratsingh@hotmail.com I have a solution 4 him.?
113997
I saw the question some minutes ago, it was written 2+2+5.\nnot 2+2=5\ntry to search it.
724027
sidharth has solved the problem correctly and given you the two values for x that satisfy the conditions of the problem.\n\nHowever, he didn't specify the POINTS on the curve.\nThey are (0,0) and (-2,-2).
207303
We could change the second + to 4 by putting a vertical small line\n\n5 + 5 + 5 = 550\n5 + 5 4 5 = 550
991565
he should have answered 5. Its the number of letters in the question\n(twelve six eight)
936497
To a mathematician 1 + 1=2 right. If I were using that equation in another context it could be interpreted differently. In a nutshell the quote is saying that people sometimes interpet the same one thing differently.
853554
Hello Rajeash\nYour additional question "\nProve that when x is even then it is impossible to write the above equation For n>2.\n"\nIsnt this what thye good o'l  Sir Fermat already has stated even for odd x ?\n\n\na^(n) - b^(n) = x^(n) \nsuppose that n is not prime, thus n = p*q, with p a prime\nthen we can write\n(a^q)^p -(b^q)^p = (x^q)^p \nso we can limit our problem to n a prime. Fermat stated his statement for p>2 thus we only have to investigate the odd primes.I do not agrede with You Rashesj that we have to investigate 4 as a separate case, because 4 = 2^2 we can write the equation as (a^2)^2 - etc , Do you agree with this ?\n\nYou also say that we can assume that if we have a solution then x will be an odd prime. If a,b,x have a common factor then we can divide that factor out. So we can assumet that they dont have a common factor. If x is even then a and b should be odd.
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Pretty slick, RED\n\nYou came on pretending that "someone else" had a problem and even gave out a wrong answer on purpose.  Clever!\n\nI can see through that, you just wanted people to give you the right answer without giving you the whole "Do your own homework" speech.\n\nNice! Score one for you.  Unless, of course, I am giving waaaay too much credit for such an ingenious plan.  In which case, x=2, not 24.
255444
jf rodrigues is correct to add 65. The others are putting in parentheses where none appear in the problem. They end up subtracting 65 instead of adding, and are getting answers that are too low by 130.\n\nThe wrong answers are obtained by people subtracting and even multiplying(!) 65 when they should be adding it.
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Nobody can help you because it's false.\n\nLet a1 = 1, a2 = 4, b1 = 2, b2 = 3, and n = 7\nThen a1 + a2 = 5, b1 + b2 = 5\nBut a1*a2 = 4, while b1*b2 = 6.\n\nYou are talking about modular arithmetic, right?
581956
The answer to your first question:\n2 x 2 x 2 = 8 (simple multiplication)\n\nThe answer to your second question:\n= 2 + 2 x 2 \n= 2 + 4 (multiplication takes precendence)\n= 6\n\nRemember the phrase "Please Excuse My Dear Aunt Sally" or the word PEMDAS.  This represents the precendence order of Parentheses, Exponents, Multiplication, Division, Addition and Subtraction.
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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.
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2^1999 x 5^2000 = \n2^1999 x 5^1999 x 5 =\n(2 x 5)^1999 x 5 =\n5 x 10^1999\nThis is 5 followed by 1999 zeroes.\nThis an exact aresult.
541116
3=2 did get a correct answer...maybe you need to review your rules of square roots.
389436
5x4+4+4+5+4/2\n=20+4+4+5+2\n=35
846794
2x/x^2 - 5x + 6 + 4/x - 2\n\nFirst we will simplify\n2x/x^2\n\nThe x on the top crosses out and the x^2 on the bottom becomes x.  We end up with 2/x.\n\n2/x - 5x + 6 + 4/x - 2\n\nNow we combine like terms.  All of the terms with just numbers will be added together.  Everything with something over x will be combined. \n\n(2/x + 4/x) - 5x + (6 - 2)\n\nWe have 6/x - 5x + 4\n\nNothing else can be combined so it is now in simplest form.
786886
Why didn't you just ask me?  You could have emailed and saved 5 points.\nMy solution did indeed contain a typographical error omitting the q from the value of m.\nAlso, I had amended the solution with a comment shortly after the question had been resolved.  (There was another small error as well.  It is very difficult sometimes typing and then proof-reading mathematical proofs in these Yahoo boxes, especially when they contain alot of special characters.)\n\nSee:\n\nhttp://answers.yahoo.com/question/index;_ylt=ApZ1XohHNUdL5xkBg8bXQEfsy6IX?qid=20061124162535AAURdJ5
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4+4 will = 1\n\nThe logic is the answers are going down from five only in odd numbers thus the answer is 1.