# Missing digit

Discussion in 'Math' started by Mark44, Jun 30, 2008.

1. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
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.

Mark

2. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
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.

3. ### thingmaker3 Retired Moderator

May 16, 2005
5,072
6
Wrong.

That's what I get for guessing.

4. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
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.

5. ### Ratch New Member

Mar 20, 2007
1,068
3
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.

(6+4)(6+4)(6+4)(6+4)(3+2)=50000==5

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

6. ### Mark44 Thread Starter Well-Known Member

Nov 26, 2007
626
1
Right you are!

7. ### Dave Retired Moderator

Nov 17, 2003
6,960
144
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!

Dave