text: 275f=527c what is the next number where taking the first digit of f and putting it last gives the c temp? i mean of course 275 degrees celsius = 527 degress farenheit using standard conversion factors. if you can answer, how did you compute the solution?  i know it is a large number, more of a math problem than a temperature one c = (5/9)(f - 32) f = (9/5)c + 32  or  f = (9/5)(c + 40) - 40 c = (5/9)(f + 40) - 40  (-40 f = -40 c)  you caught your own typo there. :-)  527 f becomes (5/9)(567) - 40 = 5*63 - 40 = 315 - 40 = 275 c 275 c becomes (9/5)(315) - 40 = 9*63 - 40 = 567 - 40 = 527 f   anyway, 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:  http://answers.yahoo.com/question/?qid=1006021800716  let f be 10^n d + [n-1 digit part called "x"] then c is 10 * [n-1 digit part called "x"] + d  so:  (5/9)(10^n d + x + 40) - 40 = 10x + d  (5/9)(10^n d + x + 40) = 10x + d + 40 10^n d + x + 40 = (9/5)(10x + d + 40) 10^n d + x + 40 = 18x + (9/5)d + 72  (10^n - (9/5))d - 32 = 17x  because 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.  (10^n - (9/5))5 - 32 = 17x 17x = 5*10^n - 9 - 32 17x = 5*10^n - 41 17x = 4999....999959  for example, with no interior 9's you have 17x = 459 yielding x = 27, giving your result 527 f = 275 c.  otherwise 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.  the first time this happens is with 527 f => 275 c, the second time is with:  17 :  49 yields 2 rem 15 17 : 159 yields 9 rem  6 17 :  69 yields 4 rem  1 17 :  19 yields 1 rem  2 17 :  29 yields 1 rem 12 17 : 129 yields 7 rem 10 17 : 109 yields 6 rem  7 17 :  79 yields 4 rem 11 17 : 119 yields 7 rem  0 17 :   9 yields 0 rem  9 17 :  99 yields 5 rem 14 17 : 149 yields 8 rem 13 17 : 139 yields 8 rem  3 17 :  39 yields 2 rem  5 17 :  59 yields 3 rem  8 17 :  89 yields 5 rem  4 <==  and 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]  as strings, where say (abc)^3 means abcabcabc then:  5 (2941176470588235)^k 27 f = (2941176470588235)^k 275 c  you 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.