A nice way to find all combinations that give a sum of N?

Question

Is there a nice way for generating a list of digits (0-9), with repetitions and a length of 6, such that the sum is N, say, 20. For example:

004673 -> 4+6+7+3=20
121673 -> 1+2+1+6+7+3=20
...

Thanks

Answer 1

['{0:06}'.format(i) for i in xrange(1000000) if sum(map(int,str(i))) == 20]

does the trick and needs about 5 seconds to return all 35127 numbers.

UPDATE - as a bonus, here comes the ugly-but-much-faster (~40 times faster) version:

result = []
for a in xrange(10):
    for b in xrange(10):
        for c in xrange(10):
            if a+b+c <= 20:
                for d in xrange(10):
                    if 2 < a+b+c+d <= 20:
                        for e in xrange(10):
                            if 10 < a+b+c+d+e <= 20:
                                f = 20 - (a+b+c+d+e)
                                result.append(''.join(map(str, [a,b,c,d,e,f])))

Answer 2

Much faster of other proposed solutions:

def iter_fun(sum, deepness, myString, Total):
    if deepness == 0:
        if sum == Total:
            print myString
    else:    
        for i in xrange(min(10, Total - sum + 1)):
            iter_fun(sum + i,deepness - 1,myString + str(i),Total) 

def fixed_sum_digits(digits, Tot):
    iter_fun(0,digits,"",Tot) 


fixed_sum_digits(6,20)

Still some room for speeder code but then the code would be boring to be read!