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        riddles >> medium >> digital root
        
(Message started by: Christine on Jan 8^th, 2013, 12:34pm)

Title: digital root
Post by Christine on Jan 8^th, 2013, 12:34pm

Why is the digital root of the product of twin primes, other than (3,5), 8?

Title: Re: digital root
Post by towr on Jan 8^th, 2013, 10:52pm

[hide]Because the digital root of n is 1+ (n-1) mod 9
And twin primes greater than (3,5) are of the form (6n-1,6n+1). So their product is 36n² - 1 = -1 (mod 9).[/hide]

Title: Re: digital root
Post by Christine on Jan 9^th, 2013, 10:47am

on 01/08/13 at 22:52:23, towr wrote:

[hide]Because the digital root of n is 1+ (n-1) mod 9
And twin primes greater than (3,5) are of the form (6n-1,6n+1). So their product is 36n² - 1 = -1 (mod 9).[/hide]

I see it now,
(6n - 1)(6n + 1) = 36*n^2 - 1
because we know that the digital root of a square is 1, 4, 7, or 9

so 36*n^2 mod 9 - 1 = -1

Title: Re: digital root
Post by towr on Jan 9^th, 2013, 11:59am

I suppose that's another way, if that's a well-known property of digital roots. It's implied by the first line of my post, but we can also apply that directly to drop the first term (since 36 is a multiple of 9).

Title: Re: digital root
Post by tsitut on Feb 20^th, 2013, 12:13pm

hm right XD I feel sh*tty I couldn't find the right answer without looking :/