Hard math problem (due to Ramanujan)?
920440
This looks like it has to be divergent and thus approach infinity, but I'm not 100%.
495174
if u meant {4/(q^2+2q+1)}- {q/7(q+1)}\nthe answer is\n\n(28-q^2-q)/{7*(1+q)^2}
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
251382
not sure this will be right but at least I'll get 2 points.\n8*8=64+8*8=64*8=1024-8-8-8=1000\n8*8=64\n64+64=128\n128*8=1024\n1024-8-8-8=1000
166778
29786\n+  850\n+  850\n_____\n31486\n\nThe way to solve this is to first note that the units digit of Y + N + N is Y. Therefore N must be 5 or 0. If N is 5 then you'd need to carry the one so that now the units digit of 1 + T + E + E is T. But this is impossible because E + E is an even number. Therefore N is 0. Using similar logic, we find E must be 5 and we do carry a one to the next column. Let's look at what we have now:\n\n___1\nFORTY\n+__T50\n+__T50\n_____\nSIXTY\n\nNow, recognize that T50 and RTY are less than 1000 and FO thousand plus two thousand (and change) can only equal SI thousand if O is 9 and I is 1. This gives:\n\n___1\nF9RTY\n+_T50\n+_T50\n_____\nS1XTY\n\nF and S must be consecutive and R and X are not both odd or both even (since we carried the one). If R is 8,=> T is 7,=> X is 3, but there remain no consecutive numbers available for F and S, so R is not 8, yet 1 + R + T + T must be at least 22, so T is therefore 8. R = 6 also fails to allow consecutive values for F and S, so R = 7.\n\n___1\nF978Y\n+_850\n+_850\n_____\nS1X8Y\n\nFrom here it should be easy to see X is 4 and F and S are 2 and 3 respectively leaving Y to be 6 and solving the puzzle.
145253
Take a look at this link. You should find all you need about Fermat Numbers there.
511948
There are an infinite number of ordered pairs that would satisfy this equation. If you want to find a few of them, assign a value to x and then solve for y.
800717
(1) First one is copied wrongly by you                                            (2) Where is the equal to sign?                                                        (3)This equn. also cannot be solved                                              (4)-p/16+5=9  Multiplying all the terms by 16 ,we get -p+80=144   =>-p=144-80=66 Therefore p=66 YOU SHOULD BE MORE CAREFUL BEFORE SENDING YOUR PROBLEMS TO YAHOO ANSWERS.HOWEVER,YOU MAY SEND THE CORRECTED VERSION OF THE PROBLEMS TO ME IMMEDIATELY THROUGH YAHOO ANSWERS.
164974
The non-brute-force solution is to notice that:\n\n   a b c a b (c+1) = (abc) * 1001 + 1\n\nSo the number needs to be a square which is 1 more than a multiple of 1001.\n\nIn number theory, we write this as:\n\n    z^2 == 1 (mod 1001)\n\nNow, we factor 1001 into primes as 1001 = 13 * 11 * 7\nand we can say that the above equation is the same as the three equations:\n\n    z^2 == 1 (mod 7)\n    z^2 == 1 (mod 11)\n    z^2 == 1 (mod 13)\n\nNow, for prime numbers, z^2 == 1 (mod p) can be solved only for z = +/- 1 (mod p)\n\nSo we get eight possible triples of equations:\n\n    z == (1,1,1)    (mod 7,11,13)\n    z == (1,1,-1)   (mod 7,11,13)\n    z == (1,-1,1)   (mod 7,11,13)\n    z == (-1,1,1)   (mod 7,11,13)\n    z == (1,-1,-1)  (mod 7,11,13)\n    z == (-1,1,-1)  (mod 7,11,13)\n    z == (-1,-1,1)  (mod 7,11,13)\n    z == (-1,-1,-1) (mod 7,11,13)\n\nNow, the equation:\n\n    z == (p,q,r)   (mod 7,11,13)\n\nis solved via the Chinese remainder theorem, and you get the following values for z:\n\n     z = 1,     z^2 = 000001\n     z = 155,   z^2 = 024025\n     z = 274,   z^2 = 075076\n     z = 428,   z^2 = 183184\n     z = 573,   z^2 = 328329\n     z = 727,   z^2 = 528529\n     z = 846,   z^2 = 715716\n     z = 1000,  z^2 = 1000000 (obviously, not a solution)\n\n[ You can think of the 1,000,000 case as when a=b=c=9. ]
823097
Can't prove it as n cannot = + or -1 and k be an interger.\nIt appears to work with other odd integers. Change your question then it may be possible to prove it.
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!
1140302
Here is an interactive page that discusses one way to do it:\n\nhttp://dilc.upd.edu.ph/loDeploy/math/math10a.html\n\nFor large numbers, bequalmings technique using logs is an easier way to go.
943730
This is Goldbach's conjecture and has NOT been proven\ntrue or false.  Good luck solving it!!\n\nI looked at the above answers, adn they dont seem to understand the question.  You shouldl have said:\n\nAny even number greater than 2 CAN be written as the sum of two primes. Prove true or false.
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}
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!
841476
When you retry the problem remember to move only yout z^-2 up don't move the 4 that is a constant in the denominator. \n\nyou should get this . (12x^5z^3/30x^7y^7) * (15x^2z^10/4y^3)\n\nremember to simplify as best you can prior to multiplying.\n\ntry from here and let me know what you get.
880382
Ok, I think that I got it. on the third one you posted, it's about simplifying the binomials (2 numbers). So you add the exponents of the like variables (or coefficiants- there's no number, then it's assumed to be 1. Ex.\n(z^4)^-3 * (z^-2)^-5\n(z^2)^-8\n\nAre there no answers in the back of the book? I would email you, but it blocks your address, if I could see some of your notes it would be easier to explain I think.