# Project Euler Problem# 5 solution in C++

/**

Project Euler problem# 5:

2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.

What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?

Below is an efficient solution of above problem using C++. Time complexity in the random access machine model is O(n log log n)

operations, a direct consequence of the fact that the prime harmonic series asymptotically approaches log log n.

*/

#include <iostream>

using namespace std;

//Finding Prime Numbers using the sieve of Eratosthenes, one of a number of prime number sieves,

//is a simple, ancient algorithm for finding all prime numbers up to any given limit.

// it is not an trial and check algorithm but find prime numbers by jumping out.

//The sieve of Eratosthenes is one of the most efficient ways to find all of the smaller primes (below 10 million or so)

int find_prime (int *numArray, int maxNum)

{

int factor = 2; // we will make 2 as the starting point.

numArray[1] = 0; // rule out 1 from our logic to avoid incorrect results.

// loop condition will check, if we are in our maximum number limit.

//maxNum is the number till there we can find the prime numbers.

while ((factor * factor) < maxNum)

{

for (int i = 2; i < maxNum; i++) // we start our itration from number 2 not from 0 or 1 to get correct results.

{

if (numArray[i] != 0) //if a number on current array index is not zero, it is a prime or we haven’t yet checked it.

{

// we are putting zeros on all multiples of prime numbers, one by one.

if (numArray[i] != factor && (numArray[i]% factor) == 0)

{

numArray[i] = 0;

}

}

}

++factor;

}// after the loop, array will have zeros on all non prime locations and prime numbers.

}

int main ()

{

int maxNum = 0;

int result = 1;

int num = 0;

cout << “enter max number: “;

cin >> maxNum;

int *myArray = new int [maxNum];

//we fill up the array till the number we want to find the smallest positive number that is evenly divisible.

for (int i = 0; i < maxNum; i++)

{

myArray[i]= i;

}

// we will get prime numbers till the maxNum by calling below function.

find_prime(myArray, maxNum);

// returning array will have prime numbers and zeros on non prime locations.

// we start our loop from 2 as first two locations are not use full so we reduce the itration of the loop.

for (int i = 2; i < maxNum; i ++)

{

// we only do our calculation on prime numbers.

if ((myArray [i]) != 0)

{

num = myArray[i];

// we take multiples of each prime number till the maxNum. meaning we raise each prime to power of a number.

// which result will remain till maxNum.

while ((num * (myArray[i])) < maxNum)

{

num = num * (myArray[i]);

}

result = num * result;

}

}// after the loop result will have smallest positive number that is evenly divisible by all of the numbers from 1 to maxNum.

cout << “smallest positive number that is evenly divisible by all of the numbers from 1 to ” << maxNum

<< ” is: ” << result <<endl;

delete []myArray;

}

###### Related articles

- Project Euler: Problem # 3 solution in C++ (alikhuram.wordpress.com)
- Largest prime number found! (quantumfrontiers.com)
- US professor finds longest prime number with 17,425,170 digits (ndtv.com)
- Prime numbers (haskell.org)

Posted on March 30, 2013, in Algorithms, C++, Programming, Project Euler and tagged Prime number, Prime numbers, Project Euler, Project Euler Problem# 5 solution in C++, Sieve of Eratosthenes, the sieve of Eratosthenes. Bookmark the permalink. 5 Comments.

Hi,

This solution does not seem to be working properly, please check.

Sorry, it works, but the answer for 1-10 must be 2520, I am getting that as the answer for 1-11.

Pingback: Project Euler Problem# 6 solution in C++ | Khuram Ali

Pingback: Project Euler Problem#7 solution in C++ (Brute Force) | Khuram Ali

Pingback: Project Euler Problem# 8 solution in C++ | Khuram Ali