The number of terminal zeros for 29! is 6, one for each '5' below 29 (25 counts for 2 factors of 5).  So e=0

To find d, eliminate 6 factors of 5 and 6 factors of 2 from the list 1,2, ... , 29 but only look at the last digit.  We'll eliminate 4 and 16 for our 6 factors of 2.  Eliminating 5 and 25 accounts for 3 factors of 5.  Then (10,15,20) become (2,3,4):

1,2,3,6,7,8,9,2,1,2,3,4,3,7,8,9,4,1,2,3,4,6,7,8,9

Running through this list multiplying but just keeping the last digit results in "6".   So, d=6

The sum of digits of the given string is 104 or if we include de=60, then 110.  The sum of all digits including the missing ones, must be a multiple of 9, e.g. 117, 126, or 135

So the sum of a+b+c is 7, 16, or 25.

Due to the divisible by 11 rule, the sum digits in the even positions minus the sum digits in the odd positions must be a multiple of 11.  Putting in 60 for de, and splitting the number accordingly gives:

odds:  84b1979094466000, sum = 67 + b

evens: a169337c5531000, sum = 43 + a + c

So 24+b-a-c must be 11, 22, or 33

Or, b-a-c must be -13, -2, or 9

b+a+c = {7,16,25}

b-a-c = {-13,-2,9}

summing the two equations, the LHS is 2b.

  For the RHS, at least at first, consider every possible pairing, but only include a sum if it is even and between 0 and 18.

  RHS can be {12, 14, 16}

So, b can be {6,7,8}

Suppose b = 6.

Then a+c = {1,10,19}

 and -(a+c) = {-19,-8,3}

 but 2 digits cannot sum to 19, so b is not 6.

Suppose b = 8

Then a+c = {-1,8,17}

 and -(a+c) = {-21,-10,1}

 so again no match.

Suppose b = 7

Then a+c = {0,9,18}

 and -(a+c) = {-20,-9,2}

 we have a match:  a+c = 9

So far, (a,b,c,d,e) = (a,7,9-a,6,0)