Missing digit

Thread Starter


Joined Nov 26, 2007
The number 2\(^{29}\) is interesting for the reason that it has nine digits in its base-10 form, and nine of the ten decimal digits are present.

Without calculating 2\(^{29}\), what's the missing digit?

The problem is trivial if you calculate the number, using a calculator or computer or by longhand multiplication.



Joined May 16, 2005
Without calculating? I can hear my father yelling at me now from 2000 miles away for making a guess instead of calculating!

Okay. I always was a disobedient child.

I'll guess "2." Somehow seems like the most "interesting" number to be missing.

I'll calculate after posting, to know if my guess is correct or not.

Thread Starter


Joined Nov 26, 2007
Here's something that might serve as a hint:

2\(^{6}\) = 64 \(\equiv\) 1 mod x
2\(^{30}\) = (2\(^{6}\))\(^{5}\) \(\equiv\) (1\(^{5}\)) mod x = 1

I'm doing modular arithmetic here, but I have not stated which modulo class I'm using. This is, after all, only a hint.


Joined Mar 20, 2007
To the Ineffable All,

Taking the hint, and using the "casting out of 9's" http://mathforum.org/library/drmath/view/55831.html .

2^29 = ((2^6)^4)*2^5=64*64*64*64*32

Applying the casting out of 9's by multiplying the sum of the digits of the multiplicand and multipliers, and summing the product digits.


Therefore the sum of the product digits of 2^29 must be 5.

The only set of digits where this happens is:

1+2+3+0+5+6+7+8+9 = 41 ==5

Therefore, 4 is the missing digit of 2^29. Ratch


Joined Nov 17, 2003
Interestingly when I read this puzzle the first thing that jumped to my mind was the cast-out of 9s method - not because I had an idea of how this worked, but because of a puzzle jpanhalt posted a while back which employed this method. Something stuck in there from that.

Good catch Ratch!