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In this article, you will learn and get code to find and print LCM and HCF (GCD) of any given two numbers by user at run-time in C++. Here are the list of programs, you will go through:

- Find LCM of Two Numbers using while Loop
- Find HCF of Two Numbers using while Loop
- Find LCM and HCF of Two Numbers in Single Program. Code used in this program is most preferred one to find LCM and HCF both
- Find LCM and HCF of Two numbers using for loop
- using Function

Before going through programs listed above, if you're not aware about, how LCM or HCF of two numbers gets calculated, then you can refer to these articles:

to clear all your doubt about LCM and HCF. Now let's move on to the program.

To find the LCF of two numbers in C++ programming, you have to ask from user to enter the two number. Then find and print its LCM on output as shown in the program given below.

**Note - **LCM is the Lowest Common Multiple or Least Common Divisor. For example, if there are two numbers say
**10** and **12**. Then its Least Common Divisor is 60. That is, both numbers **10** and **12** divides
**60** without leaving any remainder (or with leaving 0 as remainder).

// C++ Program to Find LCM of Two Numbers // ----codescracker.com---- #include<iostream> using namespace std; int main() { int numOne, numTwo, mp; cout<<"Enter Two Numbers: "; cin>>numOne>>numTwo; if(numOne>numTwo) mp = numOne; else mp = numTwo; while(1) { if((mp%numOne == 0) && (mp%numTwo == 0)) break; else mp++; } cout<<"\nLCM ("<<numOne<<", "<<numTwo<<") = "<<mp; cout<<endl; return 0; }

This program was build and run under *Code::Blocks* IDE. Here is its sample run:

Now supply any two numbers say **2** and **3**. Press `ENTER`

key to find and print its LCM
as shown in the output given below:

**Note** - **6** is the lowest number that is completely divisible by both numbers **2** and **3**.

Here is another sample run with user input, **4, 5**:

The main logic behind above program is:

- Initializes the greater number to
**mp** - Putting
**1**as condition of*while loop*always evalautes to be true. Therefore this loop ends its executing until the**break**keyword gets executed - The
**break**keyword gets executed when both the condition of**if**(inside*while loop*) evaluates to be true - Every time entering into
*while loop*'s body, a condition is applied using**if**, that checks whether the value in**mp**is divisible by both the number or not - If it is divisible, then use
**break**keyword and exit from the loop - Otherwise increment its value and continue to check with next

The dry run of above program with user input **2** and **3** goes like:

- When user enters
**2**and**3**as input. Then it gets stored in**numOne**and**numTwo**variables. So,**numOne=2**and**numTwo=3** - Now the condition of
**if**gets executed. That is, the condition**numOne>numTwo**or**2>3**evaluates to be false, therefore program flow does not goes inside**if**'s body, rather it goes to its**else**'s part, and the value of**numTwo**gets initialized to**mp**. So**mp=3** - Now the condition
**1**(of while loop), evaluates to be true, therefore program flow goes inside the loop - Inside the loop, the condition of
**if**gets evaluated. That is the condition,**mp%numOne==0**or**3%2==0**or**1==0**evaluates to be false. Because the first condition evaluates to be false, therefore no need to evaluate the second condition - So the value of
**mp**gets incremented and checks for the condition again with new value of**mp**, that is**4**for second time here - This process continues until both conditions evaluates to be true, and program flow goes inside
**if**'s body, executes the**break**keyword to stop the execution of**while loop** - When the value of
**mp**becomes**6**, then both the condition satisfies, that is**6**is divisible by both numbers say**2**and**3** - Therefore after exiting from the loop, the value of
**mp**will be 6. - Print the value of
**mp**as LCM result of given two numbers.

Unlike LCM, that deals multiple (lowest common), HCM deals with factor (highest common). HCF can be called as Highest Common Factor, or Greatest Common Factor (GCD).

For example, if there are two numbers say **10** and **12**, then its highest common factor is **2**.
That is, **2** is the highest number that divides both the number. 1 also divides both numbers, but 2 is greater,
so 2 is HCF of **10** and **12**.

The question is, **write a program in C++ that finds HCF of two numbers**. Here is its answer:

// C++ Program to Find HCF (GCD) of Two Numbers // ----codescracker.com---- #include<iostream> using namespace std; int main() { int numOne, numTwo, mp; cout<<"Enter Two Numbers: "; cin>>numOne>>numTwo; if(numOne>numTwo) mp = numOne; else mp = numTwo; while(1) { if((numOne%mp == 0) && (numTwo%mp == 0)) break; else mp--; } cout<<"\nHCF ("<<numOne<<", "<<numTwo<<") = "<<mp; cout<<endl; return 0; }

Here is its sample run with user input, **2** and **3**:

**Note - **Because **1** is the highest positive integer that divides both the number say **2** and
**3**, therefore **1** is the HCF of **2** and **3**

**Note - **Here, highest positive integer, must be less than the smaller number of given two numbers.

Here is another sample run with user input, **18** and **27**:

**Note - **Because **9** is the highest positive integer that divides both the number say **18** and
**27**

The logic used here is little similar to previous program on LCM of two numbers. The difference is in the
condition and the updatation of **mp** variable. That is, instead of incrementing it, decrement its value every
time entering into the loop. And the condition gets changed with:

if((numOne%mp == 0) && (numTwo%mp == 0))

This C++ program receives two numbers from user and finds both LCM and HCF.

// Find LCM and HCF in Single Program // ----codescracker.com---- #include<iostream> using namespace std; int main() { int numOne, numTwo, lcm, hcf, a, b, temp; cout<<"Enter Two Numbers: "; cin>>numOne>>numTwo; a = numOne; b = numTwo; while(b!=0) { temp = b; b = a%b; a = temp; } hcf = a; lcm = (numOne*numTwo)/hcf; cout<<"\nLCM ("<<numOne<<", "<<numTwo<<") = "<<lcm; cout<<"\nHCF ("<<numOne<<", "<<numTwo<<") = "<<hcf; cout<<endl; return 0; }

Here is its sample run with user input, **9, 8**:

Now let's create the same program using **for loop**

// Find LCM and HCF of Two Numbers using for Loop // ----codescracker.com---- #include<iostream> using namespace std; int main() { int numOne, numTwo, mp, temp; cout<<"Enter Two Numbers: "; cin>>numOne>>numTwo; if(numOne>numTwo) mp = numOne; else mp = numTwo; temp = mp; for(;;mp++) { if((mp%numOne == 0) && (mp%numTwo == 0)) break; } cout<<"\nLCM ("<<numOne<<", "<<numTwo<<") = "<<mp; mp = temp; for(;;mp--) { if((numOne%mp == 0) && (numTwo%mp == 0)) break; } cout<<"\nHCF ("<<numOne<<", "<<numTwo<<") = "<<mp; cout<<endl; return 0; }

Here is its sample run with user input, **10** and **12**:

**Note** - The initialization part (first part) and condition-checking part (second part) of **for loop**
is hidden. Leaving the condition empty, always evaluates to be, as true condition.

This program uses two user-defined functions namely, **findLCM()** and **findHCF()** to find LCM and HCF
of two entered numbers by user.

Both function receives two arguments, as first and second number, returns LCM/HCF of these two numbers passed as its argument.

// Find LCM and HCF of Two Numbers using Function // ----codescracker.com---- #include<iostream> using namespace std; int findLCM(int, int); int findHCF(int, int); int main() { int numOne, numTwo, res; cout<<"Enter Two Numbers: "; cin>>numOne>>numTwo; res = findLCM(numOne, numTwo); cout<<"\nLCM ("<<numOne<<", "<<numTwo<<") = "<<res; res = findHCF(numOne, numTwo); cout<<"\nHCF ("<<numOne<<", "<<numTwo<<") = "<<res; cout<<endl; return 0; } int findLCM(int a, int b) { int lcm; if(a>b) lcm = a; else lcm = b; while(1) { if((lcm%a == 0) && (lcm%b == 0)) break; lcm++; } return lcm; } int findHCF(int a, int b) { int hcf; if(a>b) hcf = a; else hcf = b; while(1) { if((a%hcf == 0) && (b%hcf == 0)) break; hcf--; } return hcf; }

Here is its sample run with user input, **6, 8**:

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