the number 2^29 contains exactly 9 distinct digits. Which digit is missing? (it has all the numbers but 1 number that can be: 1, 2, 3, 4, 5, 6, 7, 8, 9, and 0)
The number 2^29 contains exactly 9 distinct digits, but it is missing the digit 1.
We can see this by considering the following:
The units digit of 2^29 must be 2, because 2^1 = 2 and 2^4 = 16.
The tens digit of 2^29 must be 4, because 2^2 = 4 and 2^8 = 256.
The hundreds digit of 2^29 must be 8, because 2^3 = 8 and 2^12 = 4096.
The thousands digit of 2^29 must be 6, because 2^4 = 16 and 2^16 = 65536.
The ten thousands digit of 2^29 must be 4, because 2^5 = 32 and 2^20 = 1048576.
The hundred thousands digit of 2^29 must be 2, because 2^6 = 64 and 2^24 = 16777216.
The millions digit of 2^29 must be 0, because 2^7 = 128 and 2^28 = 268435456.
This leaves us with only two digits left to fill in: the billions digit and the tens of billions digit. Since the number 2^29 must have 9 distinct digits, the missing digit must be 1.
Therefore, the number 2^29 contains all of the digits 0, 2, 3, 4, 5, 6, 7, 8, and 9, except for the digit 1.