One possible solution for problem 5 on the Project Euler website

ProblemFive.cpp

/*
 * Project Name:            Problem Five
 * Solution Name:           Problem Five
 * Original creation date:  09/07/2011
 * Edit date:               18/01/2013
 * Programmer name:         Jamie Taylor (aka "GaProgMan")
 * File name:               ProblemFive.cpp
 *
 * Purpose of the project:
 * This code is my solution for "Problem Five" listed on
 * Project Euler.
 *  The original problem is listed in the comment block
 * below.
 *  Initially, my idea was to brute force my way to the
 * correct answer. The method for this is start with
 * n=1, incrementing through a loop, and each time
 * divide the number by all m numbers (1 to 20). The
 * first number to pass each m number is the correct
 * answer.
 * 
 * Problem Five, from Project Euler.
 *   URL: http://projecteuler.net/index.php?section=problems&id=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?
 *
 *    GNU Copyright information
 *    Copyright 2011 Jamie Taylor <jamie@taylorj.org.uk>
 *
 *    This program is free software; you can redistribute
 *        it and/or modify it under the terms of the GNU General
 *        Public License as published by the Free Software
 *        Foundation; either version 2 of the License, or (at
 *	your option) any later version.
 *
 *	This program is distributed in the hope that it will
 *	be useful, but WITHOUT ANY WARRANTY; without even the
 *	implied warranty of MERCHANTABILITY or FITNESS FOR A
 *	PARTICULAR PURPOSE.  See the GNU General Public
 *	License for more details.
 *
 *	You should have received a copy of the GNU General
 *	Public License along with this program; if not, write
 *	to the Free Software Foundation, Inc., 51 Franklin
 *	Street, Fifth Floor, Boston, MA 02110-1301, USA.
*/

#include <iostream>

using namespace std;

bool BruteForceCheck(long result);
bool SmarterCheck(long result);

//brute force checker
bool BruteForceCheck ( long result ) {
  for (int m = 1; m <= 20; m++){
    if (result % m != 0)
	  return false;
  }
  return true;
}

/*
 * Smarter checker
 *   Algorithm/Ideas:
 *     We are told, in the problem statement, that: "2520
 *   is the smallest number that can be divided by each
 *   of the numbers from 1 to 10 without any remainder"
 *     Since we're looking at the numbers between 1 and
 *   20, and we've been told the value for the numbers
 *   between 1 and 10, we can use that value and skip
 *   straight to checking 11 to 20.
 * Test results:
 *     On my machine, the brute force check takes 10-13
 *   seconds to complete. By using a little savvy, and
 *   moving past the first 10 steps in the check (as
 *   we're provided with the answer for the first 10),
 *   this cuts execution time down to >0.5 seconds
 */


bool SmarterCheck (long result){
  for (int m = 11; m <= 20; m++){
    if (result % m != 0)
	  return false;
  }
  return true;
}

int main () {
  long n = 1;
  /*
   * n will store the answer for this particular problem
   * m will be each of the numbers (1 to 20) that n has
   * to pass
   */
  cout << "BRUTE FORCE CHECK" << endl;
  //used for the bruteforce checker
  for (n; !BruteForceCheck(n); n++);
  cout << "The smallest number that can be evenly divided by 1 to 20 is:"
    << n << endl;

  cout << "SMARTER CHECK" << endl;
  n = 2520;
  //used for the smarter checker
  for (n; !SmarterCheck(n); n+=2520);

  cout << "The smallest number that can be evenly divided by 1 to 20 is:"
    << n << endl;
  char ch;
  cin >> ch;
  return 0;
}