
# Even Fibonacci Numbers Problem & Solution

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.

See the [even Fibonacci numbers problem on Project Euler](https://projecteuler.net/problem=2).

## C++ Solution

```cpp
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")

#include <iostream>

using namespace std;

static const int _=[](){ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);return 0;}();

int main() {
  int sum = 0;

  int first = 1;
  int second = 2;
  while (second < 4000000) {
    if (second % 2 == 0) {
      sum += second;
    }

    int tmp = first + second;
    first = second;
    second = tmp;
  }

  cout << sum << endl;
}
```


Solution for 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.

Even Fibonacci Numbers

Project Euler Even Fibonacci Numbers Solution


# Largest Palindrome Product Problem & Solution

A palindromic number reads the same both ways.
The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 * 99.
Find the largest palindrome made from the product of two 3-digit numbers.

See the [largest palindrome product problem on Project Euler](https://projecteuler.net/problem=4).

## C++ Solution

The runtime complexity of this algorithm is $O(n^2)$.

```cpp
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")

#include <iostream>

using namespace std;

static const int _=[](){ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);return 0;}();

bool is_palindrome(int n) {
  int m = 0;
  int tmp = n;
  while (tmp > 0) {
    m = m * 10 + tmp % 10;
    tmp /= 10;
  }
  
  return m == n;
}

int largest_palindrome() {
  int result = 0;

  for (int i = 999; i >= 100; --i) {
    for (int j = i; j >= 100; --j) {
      int product = i * j;
      if (is_palindrome(product)) {
        result = max(result, product);
      }
    }
  }
  
  return result;
}

int main() {
  cout << largest_palindrome() << endl;
}
```


Solution for the largest palindrome product problem from Project Euler.

Largest Palindrome Product

Project Euler Largest Palindrome Product Solution


# Largest Prime Factor Problem & Solution

The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143?

See the [largest prime factor problem on Project Euler](https://projecteuler.net/problem=3).

## C++ Solution

```cpp
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")

#include <iostream>

using namespace std;

static const int _=[](){ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);return 0;}();

int main() {
  long n = 600851475143;

  int result = 0;
  int i = 2;
  while (n > 1) {
    while (n % i == 0) {
      n /= i;
      result = max(result, i);
    }

    ++i;
  }
  
  cout << result << endl;
}
```


Solution for the largest prime factor problem from Project Euler.

Largest Prime Factor

Project Euler Largest Prime Factor Solution


# Multiples of 3 and 5 Problem & Solution

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.

See the [multiples of 3 and 5 problem on Project Euler](https://projecteuler.net/problem=1).

## C++ Solution

```cpp
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")

#include <iostream>
#include <vector>

using namespace std;

static const int _=[](){ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);return 0;}();

int main() {
  int sum = 0;
  for (int i = 0; i < 1000; ++i) {
    if (i % 3 == 0 || i % 5 == 0) {
      sum += i;
    }
  }
  
  cout << sum << endl;
}
```


Solution for the multiples of 3 and 5 problem from Project Euler.

Multiples of 3 and 5

Project Euler Multiples of 3 and 5 Solution


# Smallest Multiple Problem & Solution

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?

See the [smallest multiple problem on Project Euler](https://projecteuler.net/problem=5).

## C++ Solution

```cpp
#pragma GCC optimize("Ofast")

#include <iostream>

using namespace std;

static const int _=[](){ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);return 0;}();

int main() {

  // You can do this problem by hand; there is no point in writing an algorithm.
  cout << 19 * 17 * 16 * 13 * 11 * 9 * 7 * 5 << endl;
}
```


Solution for the smallest multiple problem from Project Euler.

Project Euler

Problems and Solutions