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        riddles >> easy >> Powers of 2
        
(Message started by: NickH on Aug 28^th, 2004, 11:45am)

Title: Powers of 2
Post by NickH on Aug 28^th, 2004, 11:45am

Does there exist an integral power of 2 such that it is possible to rearrange the digits giving another power of 2?

Title: Re: Powers of 2
Post by Barukh on Aug 29^th, 2004, 1:48am

[smiley=blacksquare.gif]
~~No: If A = 2ⁿ > A’ = 2^k, then 9|(A-A’), but A-A’ = 2^k(2^n-k-1), and n-k [le] 3.~~
[smiley=blacksquare.gif]

That's not correct...

Title: Re: Powers of 2
Post by THUDandBLUNDER on Aug 29^th, 2004, 4:10am

Let n₁ = 2^n+r
Let n₂ = 2ⁿ

And 2^n+r - 2ⁿ = 2ⁿ(2^r - 1) = 9k if we are using decimal.
So 9 must divide 2^r - 1

But, if n₁ and n₂ have the same number of digits, then r = 1, 2, or 3

Hence no such decimal number exists.

Title: Re: Powers of 2
Post by NickH on Aug 29^th, 2004, 5:36am

That's the method I used, T&B.

What if leading zeroes are allowed in the (decimal) numbers?

Title: Re: Powers of 2
Post by Barukh on Aug 29^th, 2004, 5:38am

on 08/29/04 at 05:36:23, NickH wrote:

That's the method I used, T&B.

What if leading zeroes are allowed in the (decimal) numbers?

Hmm, I used essentially the same method, but then realized that it may not work because of the leading zeros ;D

Title: Re: Powers of 2
Post by Disoriented on Aug 31^st, 2004, 5:03pm

on 08/29/04 at 04:10:34, THUDandBLUNDER wrote:

Let n₁ = 2^n+r
Let n₂ = 2ⁿ
And 2^n+r - 2ⁿ = 2ⁿ(2^r - 1) = 9k if we are using decimal.
So 9 must divide 2^r - 1

But, if n₁ and n₂ have the same number of digits, then r = 1, 2, or 3

Hence no such decimal number exists.

I lost you on line 3. Could you explain in a way that makes sense to non-math majors?
Thanks :-[

Title: Re: Powers of 2
Post by asterix on Aug 31^st, 2004, 9:23pm

From one non-math major to another, I never would have figured out the solution, but I think I can decode the answer. First, if you take any number and rearrange the digits the difference between the two will be divisible by 9. (for example, with the 2 digits a and b, ab-ba=(10a+b)-(10b+a)=9a-9b, which must be divisible by 9).
So 2^n+r=2ⁿ*2^r so the difference between our two numbers will be 2ⁿ*(2^r-1).
It must be divisible by 9, but since 2ⁿ is never divisible by 9, then 2^r-1 has to be.
Did I get that right?

Title: Re: Powers of 2
Post by THUDandBLUNDER on Aug 31^st, 2004, 10:42pm

Quote:

Did I get that right?

Yep.