# Problem 15

### 13 comments on “Problem 15”

• #include
int main()
{
int a,b,c,hcf;
scanf(“%d %d %d”,&a,&b,&c);
hcf=(a<b)?(a<c)?a:c:(b<c)?b:c;
while(1)
{
if(a%hcf==0 && b%hcf==0 && c%hcf==0)
break;
hcf–;
}
printf("%d",hcf);
return 0;
} 0
• HelpPrepInsta 0
• Bathula

#python Program

from fractions import gcd
def LCMofArray(a):
lcm = a
for i in range(1,len(a)):
lcm = gcd(lcm, a[i])
return lcm

arr1 = [120,60,40,20]
print(“LCM of arr1 elements:”, LCMofArray(arr1)) 0
• HelpPrepInsta

Thanks for contributing the code Bathula 0
• Bathula

#python

from fractions import gcd
def LCMofArray(a):
lcm = a
for i in range(1,len(a)):
lcm = gcd(lcm, a[i])
return lcm

arr1 = [120,60,40,20]
print(“LCM of arr1 elements:”, LCMofArray(arr1)) 0
• HelpPrepInsta

Thanks for contributing the code Bathula 0
• Tirtha

SOLUTION IN PYTHON

l = []
n = int(input(‘Enter the size of array of which you want to find GCD: ‘))

for i in range(n):
x = int(input())
l.append(x)

x = l
y = l

def GCD(x,y):

if x<y:
x, y = y, x

while y!=0:
x,y = y, x%y

return x

gcd = GCD(x,y) #finding gcd of first 2 numbers

for i in range(2, len(l)):
gcd = GCD(gcd, l[i])

print(gcd) 1
• Vaibhav Jain

Thanks for contributing the code Tirtha 0
• Mohd Saif

n1,n2,n3=map(int,input().split(“,”))
if(n1<n2 and n2<n3 and n1<n3):
x=n1
elif(n1n3 and n1n2 and n2<n3 and n1<n3):
x=n2
elif(n1n3 and n1>n3):
x=n3
elif(n1>n2 and n2n3):
x=n2
elif(n1>n2 and n2>n3 and n1>n3):
x=n3
for i in range(1,x+1):
if(n1%i==0 and n2%i==0 and n3%i==0):
gcd=i
print(f”GCD for {n1},{n2},{n3} is:{gcd}”) 1
• Mohd Saif

#python

n1,n2,n3=map(int,input().split(“,”))
if(n1<n2 and n2<n3 and n1<n3):
x=n1
elif(n1n3 and n1n2 and n2<n3 and n1<n3):
x=n2
elif(n1n3 and n1>n3):
x=n3
elif(n1>n2 and n2n3):
x=n2
elif(n1>n2 and n2>n3 and n1>n3):
x=n3
for i in range(1,x+1):
if(n1%i==0 and n2%i==0 and n3%i==0):
gcd=i
print(f”GCD for {n1},{n2},{n3} is:{gcd}”) 0
• sai

#include
int main()
{
int n1, n2,n3,i, gcd;
printf(“Enter two integers: “);
scanf(“%d %d %d”, &n1, &n2,&n3);
for(i=1; i <= n1 && i <= n2 && i<=n3; ++i)
{
// Checks if i is factor of both integers
if(n1%i==0 && n2%i==0 && n3%i==0)
gcd = i;
}
printf("G.C.D of %d and %d is %d %d", n1, n2,n3,gcd);
return 0;
} 0
• sai

,n3,i, gcd;
printf(“Enter two integers: “);
scanf(“%d %d %d”, &n1, &n2,&n3);
for(i=1; i <= n1 && i <= n2 && i<=n3; ++i)
{
// Checks if i is factor of both integers
if(n1%i==0 && n2%i==0 && n3%i==0)
gcd = i;
}
printf("G.C.D of %d and %d is %d %d", n1, n2,n3,gcd);
return 0;
}

its simple way to gets GCD……////// 1
• Ayantika

#include

int main()
{
int a,b,c, gcd;
scanf(“%d %d %d”,&a,&b,&c);
for(int i=1; i <=a && i <=b && i<=c; ++i)
{
// Checks if i is factor of both integers
if(a%i==0 && b%i==0 && c%i==0)
gcd = i;
}
printf("G.C.D is %d",gcd);
return 0;
} 0