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# LCM of Three Numbers C Program asked in Wipro NLTH

## Ques: How to Find LCM of three numbers.

You will find code in these languages:-

• C
• C++
• Java

If you can write a program to find LCM of two numbers. Then you can very easily write a program to find LCM of three numbers also , cause  lcm(a,b,c)=lcm(a,lcm(b,c))

“>

lcm(a,b,c)lcm(a,lcm(b,c))lcm(a,b,c)=lcm(a,lcm(b,c))

.

So here’s your c program to find LCM of three numbers.

#includeint lcm(int,int);int main(){int a,b,c,l,k;printf(“Enter any three positive integers “);scanf(“%d%d%d”,&a,&b,&c);if(a<b)l = lcm(a,b);elsel = lcm(b,a);if(l>c)k= lcm(l,c);elsek= lcm(c,l);printf(“LCM of two integers is %d”,k);return 0;}int lcm(int a,int b){int temp = a;while(1){if(temp % b == 0 && temp % a == 0)break;temp++;}return temp;}
#includeusing namespace std;int lcm(int, int, int);int hcf(int, int, int);int main(){int a,b,c;int LCM, HCF;cout<<“Enter 1st number: “;cin>>a;cout<<“Enter 2nd number: “;cin>>b;cout<<“Enter 3rd number: “;cin>>c;LCM = lcm(a,b,c);HCF = hcf(a,b,c);cout<<“LCM of “<<a<<“,”<<b<<“,”<<c<<” is = “<<LCM<<endl;cout<<“HCF of “<<a<<“,”<<b<<“,”<<c<<” is = “<<HCF<<endl;return 0;}int lcm(int x,int y, int z){long max,lcom, count, flag=0;if(x>=y&&x>=z)max=x;else if(y>=x&&y>=z)max=y;else if(z>=x&&z>=y)max=z;for(count=1;flag==0;count++){lcom=max*count;if(lcom%x==0 && lcom%y==0 && lcom%z==0){flag=1;}}return lcom;}int hcf(int p, int q, int r){int gcf=1,flag=0, count;for(count=1; flag==0;count++){if(p%count==0&&q%count==0&&r%count==0)gcf=count;if(count>p&&count>q&&count>r){flag=1;}}return gcf;}
public class LCM {public static long lcm(int[] numbers) {long lcm = 1;int divisor = 2;while (true) {int cnt = 0;boolean divisible = false;for (int i = 0; i < numbers.length; i++) {/*** lcm (n1,n2,… 0)=0.For negative number we convert into* positive and calculate lcm.*/if (numbers[i] == 0) {return 0;} else if (numbers[i] < 0) {numbers[i] = numbers[i] * (-1);}if (numbers[i] == 1) {cnt++;}/*** divide numbers by devisor if complete division i.e. without* remainder then replace number with quotient; used for find* next factor*/if (numbers[i] % divisor == 0) {divisible = true;numbers[i] = numbers[i] / divisor;}}/*** If divisor able to completely divide any number from array* multiply with lcm and store into lcm and continue to same divisor* for next factor finding. else increment divisor*/if (divisible) {lcm = lcm * divisor;} else {divisor++;}/*** Check if all numbers is 1 indicate we found all factors and* terminate while loop.*/if (cnt == numbers.length) {return lcm;}}}public static int lcm2(int num1, int num2) {if(num1==0 || num2==0){return 0;}else if(num1<0){num1=num1*(-1);}else if(num2<0){num2=num2*(-1);}int m = num1;int n = num2;while (num1 != num2) {if (num1 < num2) {num1 = num1 + m;} else {num2 = num2 + n;}}return num1;}public static void main(String[] args) {int[] numbers = {140, 72, 130};System.out.println(“*** Least Common Multiple ***”);System.out.println(“LCM(Least Common Multiple) of N numbers using Table method “);System.out.println(lcm(numbers));System.out.println(“LCM of two numbers using repetative addition”);System.out.println(lcm2(1, 72));}}

Code is not available yet, comment down the code below the page in the comment section using the same logic or different approach you want to.

### 2 comments on “LCM of Three Numbers C Program asked in Wipro NLTH”

• Varshitha

#Python
import math
def lcm(x,y,z):
lcm=x*y//math.gcd(x,y)
lcm=lcm*z//math.gcd(lcm,z)
return lcm
x=int(input())
y=int(input())
z=int(input())
print(lcm(x,y,z)) 0
• Rekha Reddy

#in python
def lcm(x,y):
if x > y:
grt=x
else:
grt=y
while(True):
if((grt%x==0) and(grt%y==0)):
lcm=grt
break
grt+=1
return lcm
n= int(input())
m=int(input())
o=int(input())
if n<m:
l=lcm(n,m)
else:
l=lcm(m,n)
if l<o:
k=lcm(l,o)
else:
k=lcm(o,l)
print(k) 0