1 / 768 = 0.00130208(3) t=1%n, for i in xrange(1,n): For your first question, why I don’t multiply by 10? Those who are familiar with mathematical notation will recognize. As a complement to Project Euler, Problem 26. Console.ReadLine(); By the way, if your program could produce the 982 numbers of the decimal periodic fraction, it would be much appreciated. } Now we have not only solved the problem, but added tests and benchmarks. I could not help further with the mathematical knowledge… but I’m sure one day, I will. public static int Totient(int n) { Longest « mixed » periodic decimal fraction : if (prevP != p) { prevP = p; totient -= (totient / p); } #print t. return maximum_r 5 digits after the decimal point and before the beginning of the period. Change ). i--; 179:{5586592178770949720670391061452513966480446927374301675977653631284916201117318435754189944134078212290502793296089385474860335195530726256983240223463687150837988826815642458100}180 I’m wondering why in your answer above your recurring cycle is 982? For 1/7 why and when do you mutiple by 10? I am definitely learning a lot here. http://img11.hostingpics.net/pics/448499pe26tableone.jpg, Table 1 shows the divisor(d)generating the longest recurring decimal fraction: Table 1 A benchmark function takes a *testing.B and iterates up to b.N, which is a number defined by the package and dependent on the stability of the timing of your function call, and produces output telling you how many times the function was run in order to perform the benchmark provide the required execution time per call. for d in range(1000, 1, -1): In other words, all the multiples of 15 are double counted because we need them to be summed as a multiple of 5, or as a multiple of 3, but not both. I am new to c# and I don’t fully grasp the concept of having that line. Number of digits in the period : 42, Interesting case : (« mixed » periodic decimal fraction) : } However, this solution is still linear in the number of elements we have to check. With the program written by Kristian (Problem 26), However, I found one possible approach. Just an observation, and I apologize if someone above already beat me to it. 1 / 12 = 0.08(3) should be for i in xrange(1,1000): for i in d: 167:{5988023952095808383233532934131736526946107784431137724550898203592814371257485029940119760479041916167664670658682634730538922155688622754491017964071856287425149700}168 Why is this not working? 113:{8849557522123893805309734513274336283185840707964601769911504424778761061946902654867256637168141592920353982300}114 I think I can add a bit more insight into where your method with multiplying by 10 and taking remainders comes from. 1 / 48 = 0.0208(3) I am curious, how did you know how to solve this one? The problem is short and easy to understand: If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. A simple brute-force approach to this is simply to iterate through all of the numbers from 1 to 999 (since we are only to check numbers below 1000), check if the number is divisible by 3 or 5, and sum the ones which are. }. remainder %= n if d-1 longest_recurring_cycle: Multiply the mixed decimal by the correct power of ten to get a simple one, then apply the same concept. What is the biggest “simple” periodic decimal fraction This is true for a simple repeating decimal of any length, so 0.123123… = 123/999, 0.121212… = 12/99, etc (the number of digits in the denominator matches the number in the numerator). We’re looking at fractions 1/d, so this becomes finding the smallest n > 0 such that there exists x with 1/d = x/(10^n – 1), or with xd = 10^n – 1.
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