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        riddles >> medium >> Number with prime factors 3 & 7, ending in 11
        
(Message started by: gkwal on May 3^rd, 2007, 11:35am)

Title: Number with prime factors 3 & 7, ending in 11
Post by gkwal on May 3^rd, 2007, 11:35am

Show that no positive integer exists whose prime factors are at most 3 and 7, and which ends in the digits 11.

Title: Re: Number with prime factors 3 & 7, ending in
Post by Grimbal on May 3^rd, 2007, 1:42pm

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What we are looking is a
N = 2^a·3^b·5^c·7^d
That ends in 11.

To end in 1, a and c must be zero.
=> N = 3^b·7^d

Let's consider it modulo 20. We want to find such an N that equals 11 (mod 20).

3³ = 27 = 7 (mod 20)
so 7^d = 3^3d
=> N = 3^b+3d (mod 20)

But the powers of 3 (mod 20) are 1, 3 ,9, 7, 1, ...
so there is no way to get 11.
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