(in this proof treat the equal sign as the congruent sign)
a number consisting of (p1) 1's is given by
(10^(p1)1)/9
say that this is congruent to r mod p
now since p is prime and p>5 then we have
GCD(9,p)=1
thus 9 has a multiplicative inverse mod p, call this inverse i
thus in the equation we can get
(10^(p1)1)/9 = r mod p
i*(10^(p1)1)/9=r*i mod p
10^(p1)1=r*i mod p
now from eulers formula since GCD(10,p)=1 we have
10^(p1) = 0 mod p thus
r*i = 0 mod p
since i is not congruent to 0 mod p then we have
r = 0 mod p
thus p divides (10^(p1)1)/9 for all primes p>5

Posted by Daniel
on 20140328 01:07:56 