A solution to the 33rd Project Euler problem

Euler 33

#Define a function to return true if we have a digit cancelling fraction
 
"""The function will take 2 inputs: [a,b],[c,d].
It will return true if ab/cd is a digit cancelling fraction."""
 
#Define a function to return a common element of x,y
 
def commonElt(x,y):
    for z in x:
        if z in y:
            return z
    else:
        return False
 
def dcfTest(x,y):
    if commonElt(x,y)==False:
        return False
    else:
        c=commonElt(x,y)
        w=int(str(x[0])+str(x[1]))
        z=int(str(y[0])+str(y[1]))
#    w/z is our fraction.
        x.remove(c)
        y.remove(c)
        if w*y[0]==z*x[0]:
            return True
        else:
            return False
            
for a in range(1,10):
    for b in range(1,10):
        for c in range(1,10):
            for d in range(1,10):
                if [a,b]!=[c,d] and dcfTest([a,b],[c,d]) and int(str(a)+str(b))<int(str(c)+str(d)):
                    print str(a)+str(b)+"/"+str(c)+str(d)+" is a dcf"
                  
                
"""This prints:
16/64 is a dcf
19/95 is a dcf
26/65 is a dcf
49/98 is a dcf """

"""answer: (1/4)*(1/5)*(2/5)*(1/2)=(1/100)"""