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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)

 Oct 28, 2023
 #1
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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.

 Oct 28, 2023

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