Prove that infinitely many numbers in the sequence
{2012, 20122012, 201220122012, 2012201220122012,....} are multiples of 101.
As defined a(1)= 2012, a(n)= a(n-1)*10000+2012
IF there is a k such that a(k)mod 101=0 then a(2k) mod 101 , a(3k) mod 101), …a(mk)mod 101 are 0 as well.
Since 2002 mod 101 =93 and 10000 mod 101=1 we can list the series' members modulo 101 as 93,85,78,71…., or recursively a(n)= a(n-1)-8 ( all modulo 101) .
Let's find for what x a(x)=0 mod 101.
a(x)=93-8*(n-1)
8*n=101
n=101,202,303,…etc
Q.E.D.
The result is obvious, since 8 and 101 share no common divisor.
Edited on April 30, 2012, 5:34 pm
Edited on April 30, 2012, 5:36 pm
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