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        riddles >> medium >> 444445-digit number and not power of 2!
        
(Message started by: gkwal on Jul 16^th, 2007, 9:27pm)

Title: 444445-digit number and not power of 2!
Post by gkwal on Jul 16^th, 2007, 9:27pm

Each of the five-digit numbers from 11111 to 99999 (both inclusive) is written on a separate card (Clearly, there are 88889 such cards). Then, the cards are arranged in an arbitrary manner to form a chain. Prove that the 444445-digit number obtained in this way (note 444445 = 88889*5) is not equal to a power of 2.

Title: Re: 444445-digit number and not power of 2!
Post by towr on Jul 17^th, 2007, 1:26am

I think it's [hide]divisible by 3[/hide]

Title: Re: 444445-digit number and not power of 2!
Post by gkwal on Jul 17^th, 2007, 8:41pm

sum of the numbers = (99999)*(99999+1)/2 - (11110)*(11111)/2

mod(sum, 3) =2

Title: Re: 444445-digit number and not power of 2!
Post by Eigenray on Jul 18^th, 2007, 12:49am

Using a calculator, [hide]1476411 < 444444 log₂(10) < 444445 log₂(10) < 1476415,[/hide] so if the number is 2^k, we can only have k = [hide]1476412, 1476413, or 1476414[/hide]. However, [hide]mod 9, the number is congruent to 11111+...+99999 = 99999*100000/2 - 11110*11111/2, which is 8 mod 9. But 2^k = 8 mod 9 only when k = 3 mod 6, which none of the possibilities are.[/hide]

Title: Re: 444445-digit number and not power of 2!
Post by towr on Jul 18^th, 2007, 1:32am

on 07/17/07 at 20:41:11, gkwal wrote:

sum of the numbers = (99999)*(99999+1)/2 - (11110)*(11111)/2

mod(sum, 3) =2

hmmm, yeah, maybe I should have actually calculated that.
I suppose that would've been too simple.

Now if we only had numbers without 0's in them (which I tacitly overlooked), then perhaps...

Also interesting, if you take every number from 111111 to 999999 and concatenate them in whatever way, the result is divisible by 21.

Title: Re: 444445-digit number and not power of 2!
Post by Eigenray on Jul 18^th, 2007, 1:56am

on 07/18/07 at 01:32:40, towr wrote:

Also interesting, if you take every number from 111111 to 999999 and concatenate them in whatever way, the result is divisible by 21.

In fact, if you concatenate the numbers from A=111...111 (n ones) to B = 999...999 (n nines), in any order, the result is divisible by A.

Edit: [hide]So it will never be a power of 2 (or 3, or 5, ...)[/hide]