“Mixed” periodic decimal fraction: Such a sequence, where the difference between each number is constant, is called a finite arithmetic progression or finite arithmetic sequence and the sum of a finite arithmetic progression is called a finite arithmetic series. 113:{8849557522123893805309734513274336283185840707964601769911504424778761061946902654867256637168141592920353982300}114 Ex. It can be used on every prime in the reverse direction to get the fastest result,, like any number divisible by 7 leaves us with 142857 or one of its permutations, I’ve done it in the third try number (since I was suspicious, I used a counter…. So I have never bothered to add something like your suggestion to avoid that problem. The reason i put the “just” in quotes, is that I am still pretty impressed by the observation which could be the key to this problem. with 0 ≤ p ≤ 10. Find the sum of all the multiples of 3 or 5 below 100. The benefit of considering the problem in this way is that the list of numbers has a closed-form solution (i.e., a formula) for calculating the sum. for (int i = 0; i currentNum) break; Simple question: I’m not clear on your solution/algorithm. 1 / 192 = 0.005208(3) 13 / 14 = 0.92857142857142857142 … } http://www.mathblog.dk/files/euler/Problem26.cs Yes, as soon as I reach a number I have already encountered before I stop. Yes, the quotient of 1/d, where d = 96 is the longest string of digits in the non-recurring part of the decimal. In the seventh calculation we get 50/7 which gives us a remainder of 1. d=dict() Is there something I’m misunderstanding? Yet this isn’t the most efficient solution. ba = BigInteger.Pow(ba, (int)(s/f)); Reducing modulo d, this becomes 0 == 10^n – 1 (mod d), or 10^n == 1 (mod d). This solution is O(n) which isn’t bad, but isn’t great either. This works fine and both produce the required answer of 233168, which we can verify with some tests in Go: And we can run the tests to ensure that everything passes and is working normally. Thanks for sharing. Fill in your details below or click an icon to log in: You are commenting using your WordPress.com account. This a common problem in combinatorics, the study of finite or countable discrete structures, and the answer is easily described by the inclusion-exclusion principle. Console.WriteLine(“Problem26”); Solution to problem 26 shows that it is generated by 1/97. However, I found one possible approach. As an example, take the decimal 0.abcdefabcdef… This can be written as abcdef/999999. It is really fast and runs in under 1ms on my computer. return totient; Nothing fancy it’s just there to handle the input of the value of n. So to solve this problem the approach is simple. This is exactly what you had above where you found the cycle length n by counting how many times you had to multiply by 10 to get a remainder of 1 (aka equal to 1 modulo the denominator). However, this solution is still linear in the number of elements we have to check. Starting with the largest prime, iterate through searching for a prime p such that 10 is a primitive root of p. Below is the C# class I wrote with the necessary methods. This code is written to handle the command line argument so that we can run the program as follows. In the fifth calculation we get 60/7 which gives us a remainder of 4. maximum_r=t So with the solution approach explained I will present the algorithm. Problem 2: Find the sum of all the even-valued terms in the Fibonacci sequence which do not exceed one million. currentNum = temp; remainder *= 10 This new number is just some power of 10 times the original one, so it will have the same recurring cycle of digits. } Thus we have found the longest recurring cycle in 1/7. I wrote my own solution it returns the write answer but 884 for the recurring cycle. ___, Sources: With that in mind, here is a deep dive into Project Euler - Problem 1. Sign up for the Mathblog newsletter, and get updates every two weeks. #calculate the first number and stor in the varibale maximum_r Ok, great. Just doing a division and then analysing the double we get out of it is not likely to give us enough digits to find the solution. index=i 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. However, I found one possible approach. 3) Microsoft Office Excel (2007). Adam Drake leads technical business transformations in global and multi-cultural environments.

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