Hi, But Gauss explained that all one needed to do was put N=100 into the formula 1/2 × (N + 1) × N resulting in the 100th number in the list without further additions. But somewhere deep down it should satisfy something in you to do this. The description of problem 1 on Project Euler reads Find the sum of all the multiples of 3 or 5 below 1000. I am afraid I can’t help you out on that one right now. linear algo. I amazed on your solution.I seen your code After i got my solution.I UNDERSTAND STILL IM IN CODE PLAY SCHOOL .I placed my time consuming code.How to see the time taken by code.it would be useful for others too. Wow!! Adding those together is almost our answer but we must first subtract the sum of every 15th natural number (3 × 5) as it is counted twice: once in the 3 summation and once again in the 5 summation. Uva online judge, code foces as well as code chef have some really interesting problems I think. An example of integer division is 10/3 = 3. This solution can handles the sum below any given number almost equally fast, if the sum can be stored in an integer. I know these are not specific things you can read, but I hope they do help you on your way. Thanks for the help. So you will need to find a way to compress the data ranges. A solution can be implemented quickly and intuitively by using an iterative approach that loops through a range of integers between 1 and 999. I will see if I can come up with some interesting reads later on. Thanks for replying… Either modulus is definitely not the way to go because it creates an O(n) i.e. The sum of these multiples is 23. Such an amazing alternate solution! five_total += five * j; Great explanation, thank you! }, I changed the upper limit to 99999 but the result by two methods came out to be different and i am wondering why? I don’t think it is that easy to answer your question. I hope I have not offended you in any way. do for (var i = 0; i < num; i++) { result+=i; I can recommend khanacademy.org which is a great learning place that will take you far. c.WriteLine(Enumerable.Range(1, 9).Sum(x=> MultiplesOf.Invoke(x, range))); int j = 0; The whole code can be written as, int result = 0; } could you please explain, I included iostream but it disappeared from the comment. With python we can express the sum like : As expected, if we calculate the sum of numbers divisible by 6 or 10, {. […] http://www.mathblog.dk/project-euler-problem-1/ (written by Kristian Edlund) C github.com/eagletmt/project-euler-c/blob/master/1-9/problem1.c […], public static void main(String[] args){ 15)? Integer divider1 = t/multiple1; The sum of numbers divisible by 6 or 10 between 1 and 999 is 124506 index++; http://www.math.brown.edu/~jhs/frint.html, um…i don’t know why no ones mentioned it yet but why not use modulo? Of course \(\left\lfloor\frac{n}{3}\right\rfloor\). See also, Project Euler 6: Sum square difference, Next » solution Project Euler Problem 2: Even Fibonacci numbers, # Single line using list comprehensions in Python, Project Euler Problem 1: Multiples of 3 and 5 Python source, Run Project Euler Problem 1 using Python on repl.it, Project Euler Problem 2: Even Fibonacci numbers. If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. In Thus the sequences for any number divisible by n can be written as n*N*(N+1)/2. { }while(index<=divider1); { if (((i % 3) == 0) || ((i % 5) == 0)) { i+=3; There are in total 100 × 101 = 10,100 beans, so each triangle must contain half this number, namely 1/2 × 10,100 = 5,050. System.out.println(“Beginning”); approx halve the iterations. But if we have numbers like 15, 30 etc, which are multiples of both 3 and 5, should we add them once or twice? Dude you are awesome! Thanks. Hope then I will turn as good as you… For a more efficient but more maths focused solution see the MathBlog post on the topic. So I think U suggested me to first study all the Programming concepts(Frm the book U suggested for efficient progrmm. where n is the divisor, and p is the greatest number we want to check. Have a nice day. } >>> for i in range(1000): Plz Pardon me for this unrelated post… […] rather than a pure math solution. The sum of numbers divisible by 6 or 10 between 1 and 999999 is 124999500006 three_total += three * i; I tried simplifying this but I’m getting the wrong number and I don’t know why. This solution is much faster than using brute force which requires loops. ((five * j) % 3 == 0)) total += i ; If the problems were small you could just make an array, but I am not sure that is a feasible approach since the N can be rather large. Func MultiplesOf = (x, range) => Why not floor(1000/3) = 333. The sum of numbers divisible by 3 or 5 between 1 and 9999 is 23331668 As the top row increases, the bottom row decreases, so the column sum always stays the same, and we’ll always have two rows and n/2 columns for any number n. If n is odd, simply start with zero instead of one. }, long long SumDivisibleBY(long int n,long int p) On the math side, I don’t think there is any one book or series of books that will lead you to be able to solve these problems. Problem 1 If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. it is even more difficult to actually recommend something without knowning your current level. >>> x = 0 , I was looking for something about number theory which is a very integral part of the Project Euler problems. Integer sum=0; long long SumDivisibleBY(long int a,long int b); for( int i=1;i<=99999;i++) while (three * i < 999); return 0; wow such an elaborate explanation…. { The iterative approach simply won’t work fast enough, but the presented closed–form will. public static void Main() int five = 5; Where is the problem? Become familiar with the notation and give your self some problems where you push yourself a bit. System.out.println(divider1+” “+divider2); We additionally pay for variant types and along with type of the books to browse. {1-10000} [23331668] The first advice here, is to have fun. Happy coding!!! sum=sum+index5; To calculate the Nth triangular number you add the first N numbers: 1 + 2 + 3 + … + N. If you want to find the 100th triangular number, you begin the long and laborious addition of the first 100 numbers. Solution took 0 ms, ##the answer will be 234168, not 233168 as given here, Solution to Project Euler, Problem 1, using Python (v.3.6.1), >>> import time As you can see, for such small problems, it takes less than a millisecond on my computer to solve, so there really is not need to find faster solutions. The sum of numbers divisible by 3 or 5 between 1 and 999 is 233168 And the output of the algorithm is once again. I just want to ask where should I start from,(Confused)…As there’s lot just then programming U hv to learn Data,Ada etc then U must knw Maths… Extended to solve all test cases for Project Euler Problem 1. }. i++; Tho you could also refactor that to a single fx and then do a yield return. Have you read this post http://www.mathblog.dk/project-euler-prolog/ as it gives you are little background for the the pieces of code you have to wrap around the functions I provide here in order to run. namespace MapReduce Sorry, I am a beginner in programming but when I compile and run the code you provided there is no result, actually if I put a printf to display the result, it shows 1000. Project Euler - Problem 1 Problem #1 If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. Etc…, Sorry but i can not understand why we subtract … >>> print(“— %s seconds —” % (time.time() – start_time)) I have started solving problems on code-chef also. The sum of numbers divisible by 6 or 10 between 1 and 99 is 1206 If it is possible, could you please take a look and share what your approach could have been? Would you pleas post the entire compilable C# source for the optimized solution ? Feel free to make comments and correct mistakes. Reduces the iteration from 1000 times to (333 +200) = 533. i.e. However, programming is more than the language, there is a whole lot to learn about algorithms and data structures which is almost generic regardless of the language you program in. int three = 3; — 0.01000356674194336 seconds —, Digits distribution pattern in the sums of multiples of 3 and 5, Ex: System.out.println(“Sum : “+sum); Find the sum of all the multiples of 3 or 5 below 1000. I would be grateful if you do. HackerRank increases the upper bound from 1,000 to 1 billion and runs 10,000 test cases. console.log(total). Since we are looping through the all the numbers, and a two comparisons and possibly one addition is needed for each number, the algorithm scales linearly with p , which in Big O notation can be expressed as O(n), not a bad scalability as a such. OK. >>> print(x) while ( i < 1000) >>> >>> }. A more general explanation on arithmetic progression is given on wikipedia. One of my personal favorites for that topic is this one … Excuse me, but how it can be that N=p/n when N=995, p=999, and n=5? sum([i for i in range(1000) if (i%3)*(i%5)==0]), // A Map/Reduce pattern to solve this problem. 2 Now that the fluff around the coding is covered, we are ready to solve the first problem. N is the highest number less than p divisible by n. In the case of 5 this is 995. A formula attributed to Carl Friedrich Gauss will calculate the sum of the first n natural numbers.

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