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.
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 = 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;
}// 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++)
// we will get prime numbers till the maxNum by calling below function.
// 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;
- 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.