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 Many Multiples (Posted on 2012-04-30)
Prove that infinitely many numbers in the sequence {2012, 20122012, 201220122012, 2012201220122012,....} are multiples of 101.

 No Solution Yet Submitted by K Sengupta Rating: 4.0000 (2 votes)

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 spoiler | Comment 3 of 8 |

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
 Posted by Ady TZIDON on 2012-04-30 15:31:56

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