# LCM of a Number using Recursion in Python

## LCM of a Number using Recursion

On this page we will learn to create a python program to find LCM of a Number using Recursion.

LCM – Lowest common multiple of two or more number. Is Smallest number that it is completely divisible by all the numbers for which we are finding LCM.

Example :

• Input : first = 23, second = 69
• Output : HCF of 23 and 69 is 69
• Explanation : No other number less then 69 can be divide by both 23 and 69 completely. That’s why 69 is LCM of 23 & 69 ### Algorithm

• Start by making a function and passing both number to it as a and b
• Return a multiplied divided by the value returned by another function which takes a and b
• If b is equals to zero return a
• Else return recursive call for the function with values b and remainder when a is divided by b respectively ### Python Code

Run
```def hcf(a, b):
if b == 0:
return a
else:
return hcf(b, a % b)

def lcm(a, b):
return (a * b) // hcf(a, b)

first = 23
second = 69

print("Lcm of", first, "and", second, "is", lcm(first, second))```
`Output :Lcm of 23 and 69 is 69`

## Algorithm

• Start by making a function and passing both number to it as a and b
• Return a multiplied by b divided by the value returned by another function which takes a and b
• If maximum between a & b is divided by minimum between a & b gives remainder zero return minimum between a & b
• Iterate using for loop between range one more then half of minimum between a & b to 0 in reverse order using variable i
• For each iteration check if  a divided by i and b divided by i both are equals to 0 then return i

### Python Code

Run
```def hcf(a, b):
if max(a, b) % min(a, b) == 0:
return min(a, b)
for i in range(1 + min(a, b) // 2, 0, -1):
if a % i == b % i == 0:
return i

def lcm(a, b):
return (a * b) // hcf(a, b)

first = 23
second = 69

print('LCM of', first, 'and', second, 'is', lcm(first, second))```
`Output :Lcm of 23 and 69 is 69`

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