text: C = (5/9)(F - 32)\nF = (9/5)C + 32\n\nor\n\nF = (9/5)(C + 40) - 40\nC = (5/9)(F + 40) - 40\n\n(-40 F = -40 C)\n\nYou caught your own typo there. :-)\n\n527 F becomes (5/9)(567) - 40 = 5*63 - 40 = 315 - 40 = 275 C\n275 C becomes (9/5)(315) - 40 = 9*63 - 40 = 567 - 40 = 527 F\n\n\nAnyway, these are fairly straightforward to TRY to solve -- there is not necessarily a solution.  Another example where there is also a solution (method not shown there, but it is essentially the method used here) is my reply to:\n\nhttp://answers.yahoo.com/question/?qid=1006021800716\n\nLet F be 10^n d + [n-1 digit part called "x"]\nthen C is 10 * [n-1 digit part called "x"] + d\n\nSo:\n\n(5/9)(10^n d + x + 40) - 40 = 10x + d\n\n(5/9)(10^n d + x + 40) = 10x + d + 40\n10^n d + x + 40 = (9/5)(10x + d + 40)\n10^n d + x + 40 = 18x + (9/5)d + 72\n\n(10^n - (9/5))d - 32 = 17x\n\nBecause of the fraction 9/5 there, we see that d must be divisible by 5: 0 or 5 -- But it is a leading digit and so it is not 0.  Therefore d = 5.\n\n(10^n - (9/5))5 - 32 = 17x\n17x = 5*10^n - 9 - 32\n17x = 5*10^n - 41\n17x = 4999....999959\n\nFor example, with no interior 9's you have 17x = 459 yielding x = 27, giving your result 527 F = 275 C.\n\nOtherwise just keep dividing 17 into 4999....  Anytime you see a remainder of 4 you can stop the nine's and append that trailing 59 and then you will see that your quotient for x ends in 27.\n\nThe first time this happens is with 527 F => 275 C, the second time is with:\n\n17 :  49 yields 2 rem 15\n17 : 159 yields 9 rem  6\n17 :  69 yields 4 rem  1\n17 :  19 yields 1 rem  2\n17 :  29 yields 1 rem 12\n17 : 129 yields 7 rem 10\n17 : 109 yields 6 rem  7\n17 :  79 yields 4 rem 11\n17 : 119 yields 7 rem  0\n17 :   9 yields 0 rem  9\n17 :  99 yields 5 rem 14\n17 : 149 yields 8 rem 13\n17 : 139 yields 8 rem  3\n17 :  39 yields 2 rem  5\n17 :  59 yields 3 rem  8\n17 :  89 yields 5 rem  4 <==\n\nand you could either stop here, or the next line will be 17 : 49 again, so x can be any number (0 or more) copies of 2941176470588235 yielding 5(2941176470588235)*27 as the F value and therefore (2941176470588235)*275 as the corresponding C value [the "*" means that the 2941176470588235 is repeated zero or more times]\n\nAs strings, where say (abc)^3 means abcabcabc then:\n\n5 (2941176470588235)^k 27 F = (2941176470588235)^k 275 C\n\nYou may also look up stuff about when 10 is a primitive root modulo a prime and/or about discrete logarithms modulo a not necessarily prime integer.