using c = System.Console; public class Test while ( j < 1000) i++; Why not floor(1000/3) = 333. Calculating the number of beans in this rectangle built from the two triangles was easy. The whole code can be written as, int result = 0; Then do the same for 1000/5 = 200 times. Won’t the arithmetic/geometric approach double add numbers that are both multiples of 3 and 5 (e.g. While the other students labored away, the ten–year–old Gauss handed his teacher the tablet with his answer within seconds. … — 0.01000356674194336 seconds —, Digits distribution pattern in the sums of multiples of 3 and 5, Ex: I promise I will include cool tidbits for you. And no you haven’t offended me by asking this kind of question. it is important to note the parenthesis around p/n, if these are not there, the result will deviate since the program n*below/n = n, which is different from n*(below/n). I don’t think it is that easy to answer your question. Plz Pardon me for this unrelated post… I do use codechef… var answer = range.Select(r => (x % r) == 0); 5*1+5*2+…+5*199=y For example, when n=10 the sum of all the natural numbers from 1 through 10 is: (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) = 10*11 / 2 = 55. You can see a whole lot of ideas right here http://www.mathblog.dk/programming-challenges/ © 2020 mathblog.dk. Remember that the reason I wrote this, is not to give you the answer, but rather to explain different approaches to the solution. Here’s how the adaptation works: Each column sums to 33 and, using our understanding from above, we calculate 6*33=198 to find the sum of numbers from 0 to 33 that are evenly divisible by 3. First of all the programming part, it is not necessary to know more than one programming language, but there is certainly a benefit to it I think. 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. I simply don’t have time. Thanks and sorry for bothering you. >>> for i in range(1000): Plz reply… The game of bowling, or ten–pin, sets 10 pins in a equilateral triangular form: one pin in the first row through 4 pins in the last row. 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. ((five * j) % 3 == 0)) It was from the printf I wrote wrong. return 0; A geometric explanation is given here and an arithmetic explanation is given here. I will try to bang my head on the question. But somewhere deep down it should satisfy something in you to do this. Here we use integer division, which means that we will discard the fractional part of the result. Ans=x+y-z. if(index%3!=0&&index
i == true) ? […] 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){ Iterate 333 times 1*3, 2*3, 3*3 etc. So are there any books(specific for Maths) which are prefectly meant for programming purpose from zero to high…As I find myself lame for questions of Higher level on Project Euler…and I don’t want to submit every sol.n after peeping into yours.. 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. Indeed, Gauss’s teacher liked to assign these meddlesome problems to keep his class busy and quiet. Integer index = 1; jumlah=jumlah+3; I will see if I can come up with some interesting reads later on. long long SumDivisibleBY(long int a,long int b); for( int i=1;i<=99999;i++) 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. j++; If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. do sum([i for i in range(1000) if (i%3)*(i%5)==0]), // A Map/Reduce pattern to solve this problem. Wow!! But if we have numbers like 15, 30 etc, which are multiples of both 3 and 5, should we add them once or twice? However, lets explore the problem a bit more. As the \(lcm(5,3)=15\), we have to subtract all multiples of 15, so we finally get: \(\sigma(999, 3) + \sigma(999, 5) - \sigma(999, 15)\). I hope you have found it useful and have learned from it. x % 3 == 0 The iterative approach simply won’t work fast enough, but the presented closed–form will. In our Python function, sumn() (shown below), this is accomplished by taking the floor of n divided by d to find the number of non–zero terms. if (((i % 3) == 0) || ((i % 5) == 0)) { HackerRank increases the upper bound from 1,000 to 1 billion and runs 10,000 test cases. The sum of numbers divisible by 6 or 10 between 1 and 999 is 124506 However, with integer division N can be expressed as, Now we have all the parts needed to express an efficient solution, which in C# can be written as. )and for strengthening math go from High School level to beyond…and then switch to solving. >>> x = 0 Integer divider1 = t/multiple1; This page lists all of my Project Euler solution code, along with other helpful information like benchmark timings and my overall thoughts on the nature of math and programming in Project Euler. int sum = 0; sum(unique([3:3:999,5:5:999])). console.log(total). sum=sum+index3; System.out.print(index); um… That is exactly what I do in my first solution. Hi, My first suggestion to solving one of these problems, is usually to bruteforce it. The sum of numbers divisible by 6 or 10 between 1 and 999999 is 124999500006 var range = new List { 3, 5 }; if(((i%3)==0||(i%5)==0)) can you please suggest me other sites where i can practice c pragramming while (five * j < 995); Integer divider2 = t/multiple2; Etc…. If what you are really looking for is some programming challenges to throw yourself at, there are a lot of options which are less math focused then Project Euler, not that I will try to discourage you by any means. Project Euler 1 Solution: Multiples of 3 and 5 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. y=sum(range(0, 1000, 5)) SumDivisbleBy(3, 999) + SumDivisbleBy(5, 999) – SumDivisbleBy(15, 999), You could use Linq to do that: Enumerable.Range(1, 999).Where(item => item % 3 == 0 || item % 5 == 0).Sum(), A nice MATLAB-solution is: 15*1+15*2+…+15*66=z Now that the fluff around the coding is covered, we are ready to solve the first problem. I don’t know why it doesn’t work for you. a relatively simple pattern is obtained: The sum of numbers divisible by 6 or 10 between 1 and 9 is 6 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… }, I changed the upper limit to 99999 but the result by two methods came out to be different and i am wondering why? Either modulus is definitely not the way to go because it creates an O(n) i.e. } I take it as a compliment. long long j = n*(p/n)*((p/n)+1)/2; for(var i=1;i<=(1000/3);i++) {1-100000} [2333316668] approx halve the iterations. The sum of these multiples is 23. } }. All multiples of 3 for example are \(3, 6, 9, 12, ...\) If we place 3 outside the brackets, it reads \(3\cdot (1, 2, 3, 4, ...)\). { I have started solving problems on code-chef also. In the first bit of code we check if a number was divisible by 3 and/or 5, and this way we only checked each number once. Integer sum=0; Thanks. Integer multiple1 = 3; Find the sum of all the multiples of 3 or 5 below 1000. The program runs instantly for upper bounds like 1000, but does not scale well for larger ones such as 109. Would you pleas post the entire compilable C# source for the optimized solution ? However, we need to put something clever inside the SumDivisbleBy function, otherwise we have an even slower solution. 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. 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. Thanks for the tips…Very useful…gave me a direction to go upon. We additionally pay for variant types and along with type of the books to browse. I don’t mean you need to sit and laugh all the time, you might even be frustrated some of the time. Extended to solve all test cases for Project Euler Problem 1 HackerRank increases the upper bound from 1,000 to 1 billion and runs 10,000 test cases. Excuse me, but how it can be that N=p/n when N=995, p=999, and n=5? result += i; The sum of these multiples is 23. In general, sum the numbers less than 1000 that are divisible by 3 (3, 6, 9, 12, 15, …) or 5 (5, 10, 15, …) and subtract those divisible 3 and 5 (15, 30, 45, …). The sum of all numbers dividable by 3 or 5 is: 233168 Thanks for replying… Great explanation, thank you! while (three * i < 999); The sum of numbers divisible by 6 or 10 between 1 and 99999 is 1249950006 The summation formula is the legacy of Carl Friedrich Gauss, the German mathematician. sum+= (3*i); return j; Of course \(\left\lfloor\frac{n}{3}\right\rfloor\). The sum of the multiples of 3 or 5 can be calculated quite simple by looping from 1 to 999 and check what numbers are divisible by 3 and 5: Scanning through a small range like this is no problem for a computer. 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. But a linear complexity for this problem isn't satisfying. Console.WriteLine("The sum of multiples of 3 and 5 number is " + (Result )); http://www.mathblog.dk/project-euler-prolog/, http://www.mathblog.dk/stopwatch-a-timing-function-in-c/. Implementing it in JavaScript can then look like this: Project Euler 1 Solution: Multiples of 3 and 5. jumlah=jumlah+5; x : 0; The sum of these multiples is 23. for (int i = 1; i < 1000; i++) { 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.