Consider the mod 7 value of a 1 in different decimal positions.  There is a repeating cycle of period 6 regarding the mod 7 value of successive powers of 10.  Starting with 10^0:  [1,3,2,6,4,5].

The 29! number has 31 digits, so from left to right, the mod 7 values of a 1 in each position follows this pattern:

[1,5,4,6,2,3,1,5,4,6,2,3,1,5,4,6,2,3,1,5,4,6,2,3,1,5,4,6,2,3,1]

8a4176199373970c954543616000000

Making a "sumproduct":  1*8 + 5*a + 4*4 + 6*1 + ...

Finds a sum of 346 + 5a + 6c , or to put into mod 7

3 + 5a + 6c  which must equal 0 since 29! is divisible by 7

And we know that a+c = 9.  So c = 9-a

3 + 5a + 6(9-a) = 0 mod 7

57 - a =  0 mod 7

1 - a = 0 mod 7, so a is {1,8} and c is {8,1}

In retrospect, choosing mod 7 was not able to narrow it down enough.

Repeat the exercise for mod 17

[1, 10, 15, 14, 4, 6, 9, 5, 16, 7, 2, 3, 13, 11, 8, 12] 

when reversed is

[12, 8, 11, 13, 3, 2, 7, 16, 5, 9, 6, 4, 14, 15, 10, 1]

[8, 11, 13, 3, 2, 7, 16, 5, 9, 6, 4, 14, 15, 10, 1, 12, 8, 11, 13, 3, 2, 7, 16, 5, 9, 6, 4, 14, 15, 10, 1]

988 + 11a + 12c = 0 mod 17

2 + 11a + 12c = 0 mod 17; substitute c = 9-a

2 + 11a + 12(9-a)

110 - a

8 - a = 0 mod 17

So a=8 and c=1

(a,b,c,d,e) = (8,7,1,6,0)

29! = 8841761993739701954543616000000