## CODE

# IN C PROGRAMMING

#include <stdio.h>

int main()

{

int num1, num2, i, lcm;

printf(“Enter two numbers: “);

scanf(“%d %d”, &num1, &num2);

for (i = num1; i <= num1 * num2; i++) {

if (i % num1 == 0 && i % num2 == 0) {

lcm = i;

break;

}

}

printf(“LCM of %d and %d is %d\n”, num1, num2, lcm);

return 0;

}

## P R O G R A M I N G

#### Table of Contents

## E X P L A N A T I O N

### 1.Including the Standard I/O Library

*#include <stdio.h>*

###### IN C PROGRAMMING

### 2.The main Function

*int main() {*

*// …*

*}*

The main function is the entry point of the program. Here’s what’s happening inside the main function:

### 3. Variable Declarations

*int num1, num2, i, lcm;*

`num1`

and`num2`

are integer variables that will store the two numbers provided by the user.`i`

is an integer variable used as a loop counter.`lcm`

is an integer variable to store the calculated least common multiple.

### 4. Input from the User

*printf(“Enter two numbers: “);*

*scanf(“%d %d”, &num1, &num2);*

This part of the code prompts the user to enter two numbers. The `printf`

function displays the message “Enter two numbers: “, and the `scanf`

function reads two integers from the user and stores them in the variables `num1`

and `num2`

.

### 5. Calculating the LCM

*for (i = num1; i <= num1 * num2; i++) {*

*if (i % num1 == 0 && i % num2 == 0) {*

*lcm = i;*

*break;*

*}*

*}*

This is the core part of the program where the LCM is calculated using a `for`

loop and a `break`

statement. Let’s break it down step by step:

- The
`for`

loop initializes`i`

to the value of`num1`

. The loop continues as long as`i`

is less than or equal to the product of`num1`

and`num2`

. The loop increments`i`

by 1 in each iteration. - Inside the loop, there is an
`if`

statement that checks if`i`

is divisible by both`num1`

and`num2`

without any remainder. This is done using the modulo operator (`%`

), which calculates the remainder of the division. - If
`i`

is divisible by both`num1`

and`num2`

(i.e.,`i % num1 == 0`

and`i % num2 == 0`

), it means`i`

is a common multiple of both numbers. - When a common multiple is found, the
`lcm`

variable is assigned the value of`i`

, and the`break`

statement is executed. The`break`

statement immediately exits the loop, as there’s no need to continue searching for more common multiples. This is an optimization to avoid unnecessary iterations.

The result of the LCM calculation is stored in the `lcm`

variable.

### 6. Displaying the LCM

*printf(“LCM of %d and %d is %d\n”, num1, num2, lcm);*

`printf`

function to display the result. It shows the LCM value, along with the input numbers for clarity.

### 7. Return Statement

*return 0;*

The `return 0;`

statement is used to indicate that the program has completed successfully and is returning an exit status of 0. In C, a return status of 0 conventionally indicates a successful execution, while a non-zero status indicates an error.

### How the Code Works

Now that we’ve broken down the code into its components, let’s understand how the code works and calculate the LCM:

**Input**: The user is prompted to enter two numbers, which are stored in the variables`num1`

and`num2`

.**Loop for LCM Calculation**: A`for`

loop is used to iterate through numbers starting from`num1`

and up to the product of`num1`

and`num2`

. The loop variable`i`

is incremented in each iteration.**Checking for Common Multiple**: Within the loop, each value of`i`

is checked to see if it is a common multiple of`num1`

and`num2`

. This is done by using the modulo operator (`%`

) to verify that`i`

is divisible by both`num1`

and`num2`

without any remainder.**Finding the LCM**: When a common multiple is found, the program assigns that value to the`lcm`

variable. The`break`

statement is used to exit the loop immediately, as there is no need to continue searching for more common multiples. This optimization improves the efficiency of the code.**Displaying the LCM**: Finally, the calculated LCM is displayed to the user using the`printf`

function.

### Code Analysis

This code provides a straightforward approach to finding the LCM of two numbers. However, there are some points to consider:

**Optimization**: The code uses a`for`

loop to find the LCM, and it stops as soon as a common multiple is found using the`break`

statement. This is efficient because it doesn’t unnecessarily check all values up to`num1 * num2`

. Once the LCM is found, there is no need to continue iterating. However, in the worst-case scenario, if`num1`

and`num2`

are large, this method may require many iterations.**GCD vs. LCM**: While the code calculates the LCM directly, it doesn’t use the GCD (Greatest Common Divisor) in the process. It’s important to note that there’s a well-known relationship between LCM and GCD, which is often used in more efficient algorithms for LCM calculation. The relationship is:`LCM(a, b) = (a * b) / GCD(a, b)`

. Utilizing the GCD can lead to more efficient LCM calculation.**Error Handling**: The code assumes that the user will input valid numbers. It doesn’t handle cases where the user might enter zero or non-integer values. In practice, it’s a good idea to add input validation to ensure the program doesn’t encounter unexpected errors.**Additional Features**: This code is a basic LCM calculator. Depending on the context and requirements, you might want to expand it to handle more numbers or provide additional functionalities.- In summary, this C program calculates the LCM of two numbers using a
`for`

loop and a`break`

statement. It is a straightforward approach that is easy to understand. However, for large numbers, it may not be the most efficient method, and there are more optimized algorithms available for LCM calculation that take advantage of the GCD.

MORE

- SUM OF TWO NUMBER: https://d-techsol.in/write-a-program-of-addition-of-number-in-c/

BEST C PROGRAMMING BOOKS BUY LINK :

thank you