Integers and solution?
954164
grab a calculator and try 1.1 x = = = = =  until you get a number >2.\nthe first time you hit = is ^2, next is ^3 etc.  count.\n\nN=7
363348
Are we solving for x?
222284
duh...?  what's your question, Mam?  \n\n=0(
889893
2+7\n3+6\n9+0\n\nThat's it. They all add up to 9... I think that's what the question is asking.
531205
2 + 2 = 4\n\nthats all I got
77748
If we solve by setting n as the smallest integer, n+1 as the second, and n+2 as the largest:\nn+n+1+n+2=100.\n3n+3=100\n3n=97\nn=32 1/3, so the other two integers are 33 1/3 and 34 1/3.  The answer has been posted but I thought showing the process might be of interest...but these are all not integers so this is a trick question.
900761
Let your 2-digit number be represented by 10A + B.\n\nThen, we are given :\n(1) A + B = 6\nand\n(2) AB = (10A + B) / 3\n\nFrom (1) we get B = 6 - A\n\nSubstituting this into (2) gives :\nA(6 - A) = (10A + 6 - A) / 3\n\n6A - A^2 = (9A + 6) / 3 = 3A + 2\n\nTherefore, A^2 - 3A + 2 = 0\nFactorise to give : (A - 1)(A - 2) = 0\n\nThus A = 1 or 2.\n\nIf A = 1, then B = 5.\nIf A = 2, then B = 4\n\nThere are thus 2 solutions for AB.\nAB = 15 or 24.\n\n(It helped to see the question in words)
427860
There is an infinite number of solutions, \nthe set of solutions is :\n{(k,-k+12):k belongs to the set of real numbers}
770779
Call the integers x, x+1, x+2\n\nx+1+x+2 = x/2 + 24\n\n2x + 3 = x/2 + 24\n2x = x/2 + 21\n4x = x + 42\n3x = 42\nx = 14\n\nThe three integers are 14, 15, 16
137681
Ok...\nI see what you are doing...\nYou merely substitute the number in for x...\nso the problem becomes 6-2*(-1)\n6+2\n8
329355
Try solving this one:\n\nDetermine all pairs of positive integers (a, b) such that\n\na^2/(2ab^2-b^3+1)\n\nis a positive integer.\n\nNote that in 2ab^2 just the b is squared.\n\nGood luck!
695947
D\n\nEither is sufficient.  \n1) will lead to 2 equations,2 unknowns which yield a unique solution for x and y:  x=y+1, x+y=77, therefore x=39, y=38.  xy=1482 \n\n2) If the sum is 77 and x and y have the same tens digit, then there are two solutions for {x,y}, namely {38,39} and {39,38}, however the product xy is the same. xy=1482
601672
Okay, Lets just assume that you are dividing something by (n-3). Whether you are right or not doesn't matter, since you believe you are right, and I'm too lazy to check it out on my own.\n\nWhen n=3, your solution will be undefined. If you were trying to graph this, there would be an open circle around n=3. It would not - *NOT* - be infinite.\n\nNow, if you feel you are SUPPOSED to get an answer for every number you plug in, then I would go back and re-check your algebra, because if you *KNOW* you are supposed to get an answer, then something went awry with the algebra.
296877
It has a solution, indeed. It actually has two solutions, -3 and 7. To find it out, set each pair of nomials, (e.g. (x + 3)) to zero. So it would be (x+3)=0. Then solve for x. Do the same for the other nomial, and there you have it.
115650
--5 2 0 1 -----  F L O W\n--4 2 0 1 -----  B L O W\n--1 0 2 5 -----  W O L F\n-------------------------\n1 0 4 2 7 ----- N O B L E\n\nBut still N and W value are coming same...does this problem have solution. If you have then can you please send the answer to my mail id hssiddhu@yahoo.com
904203
Let the two integers be 'x' and 'x+2'\nx + x + 2 = 4^3\n2x + 2 = 64\n2x = 64 - 2\n2x = 62\nx = 62/2\nx = 31\nThe integers are 31 and 33.
227387
D = 36 - (4)*(-2)*(-3)= 36- 24 = 12\nTwo real solutions.