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Comprehensive Collection of C Programming Examples
In this example, you will learn how to calculate the LCM (least common multiple) of two numbers entered by the user.
To understand this example, you should understand the followingC programmingTopic:
Two integers n1and n2The LCM is the smallest positive integer that can be divided by n1and n2Divisible completely (no remainder). For example,72and12The LCM of 0 is360.
#include <stdio.h> int main() { int n1, n2, min; printf("Enter two positive integers: "); scanf("%d %d", &n1, &n2); //n1and n2The maximum value between them is stored in min min = (n1 > n2) ? n1 : n2; while (1) { if (min % n1 == 0 && min % n2 == 0) { printf("%d and %d of LCM is %d.", n1, n2, min); break; } ++min; } return 0; }
Output Result
Enter two positive integers: 72 120 72and12The LCM of 0 is360.
In this program, the integers entered by the user are stored in the variables n1and n2.
n1and n2The maximum value stored in min, and the LCM of the two numbers cannot be less than min.
in each iteration.
Check if min can be divided by n1and n2Divisible completely.
if (min % n1 == 0 && min % n2 == 0) { ... }
If this test condition is not true, then min will increment1, and continue iterating until the test expression of the If statement is true.
You can also use the following formula to find the LCM of two numbers:
LCM = (num1*num2)/GCD
Learn how to find GCD in C programmingGCD of two numbers.
#include <stdio.h> int main() { int n1, n2, i, gcd, lcm; printf("Enter two positive integers: "); scanf("%d %d", &n1, &n2); for (i = 1; i <= n1 && i <= n2; ++i) { // Check if i is a factor of two integers if (n1 % i == 0 && n2 % i == 0) gcd = i; } lcm = (n1 * n2) / gcd; printf("The LCM of two numbers %d and %d is %d.", n1, n2, lcm); return 0; }
Output Result
Enter two positive integers: 78 150 two numbers78and15The LCM of 0 is1950.