29*log(2) = 8.72987, then 2^29 has 9 digits total. KS has told us there are 9 distinct digits in its representation. So each of nine digits occurs exactly once in 2^29

The sum of all ten digits mod 9 equals 0.

The (sum of the digits of 2^29) mod 9 will equal 2^29 mod 9.

2^29 mod 9 = ((2^6)^4)*(2^5) mod 9 = (64^4)*32 mod 9 = (1^4)*5 mod 9 = 5.

5 + 4 = 0 mod 9. Therefore 4 must be the missing digit.

Checking by direct calculation 2^29 = 536870912