why type(t) is Nonetype?????

Question

def ley(x,y,gg=1):
    print(gg)
    if(gg%x==0)&(gg%y==0):
        print(type(gg))
        return gg
    else:
        gg+=1
        ley(x,y,gg)
t=ley(2,3,1)
print(type(t))

Answer 1

Bcoz your function is non returning type hence its type(t) is none simply t=ley(2,3,1) will be treated as simply ley(2,3,1) and also function is of NoneType

Answer 2

Your popping out of your function in the "else" condition.  Since you're not returning a value there you are getting the NoneType response.