C++ Exercises: Find LCM of any two numbers using HCF
C++ For Loop: Exercise-29 with Solution
Write a program in C++ to find the LCM of any two numbers using HCF.
Pictorial Presentation:

Sample Solution:-
C++ Code :
#include <iostream>
using namespace std;
int main()
{
int i, n1, n2, j, hcf = 1, lcm;
cout << "\n\n LCM of two numbers:\n";
cout << "------------------------\n";
cout << " Input 1st number for LCM: ";
cin >> n1;
cout << " Input 2nd number for LCM: ";
cin >> n2;
j = (n1 < n2) ? n1 : n2;
for (i = 1; i <= j; i++) {
if (n1 % i == 0 && n2 % i == 0) {
hcf = i;
}
}
/* mltiplication of HCF and LCM = the multiplication of these two numbers.*/
lcm = (n1 * n2) / hcf;
cout << " The LCM of " << n1 << " and " << n2 << " is: " << lcm << endl;
}
Sample Output:
LCM of two numbers: ------------------------ Input 1st number for LCM: 15 Input 2nd number for LCM: 25 The LCM of 15 and 25 is: 75
Flowchart:

C++ Code Editor:
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Previous: Write a program in C++ to find the number and sum of all integer between 100 and 200 which are divisible by 9.
Next: Write a program in C++ to display the number in reverse order.
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