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Project Euler Problems

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Project Euler Problem 1

Add all the natural numbers below one thousand that are multiples of 3 or 5. If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000. Solution # This problem calls for the sum of all natural numbers that are multiples of 3 or 5 below 1000. The smallest number which is a multiple of 3 or 5 is 3. So my starting number is 3. The terminating condition is also given in the problem as below 1000. Consider the following things while programming Check whether the number is a multiple of 3. Check if the number is multiple of 5 and not a multiple of 3.(This step is to avoid adding duplicate numbers). Finding the sum of all multiples of 3 or 5 below 20. Initially sum = 0 3 is the smallest number that is a multiple of 3 or 5. Implementation int main() { int sum, number; sum = 0; for (number = 3; number < 1000; number++) { if ((number % ...

Project Euler Problem 2

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms. Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ... By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms. Solution To solve this problem, let us see Fibonacci sequence whose values do not exceed 10. The Fibonacci sequence starts with 0 and 1. Add the previous two numbers p and c to get the new Fibonacci number. But our problem is to add only even numbers in the Fibonacci sequence that do not exceed four million, so each time a new Fibonacci number is found, check if the number is even and is not exceeding four million. Implementation int MAX_NUM = 4000000; int main() { int new_f, prev_f, curr_number, sum; new_f = 0; prev_f = 0; curr...

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