Prove that all numbers of the form 12008, 120308, 1203308, ... are divisible by 19.

aside: ah, Ady, don't you have an extra "+1" in your posting? 

This is a very clumsy proof, but here goes:

1st show:

10 x 120 = 19 x 63 + 3

10 x 1203 = 19 x 633 + 3

10 x 12033 = 19 x 6333 + 3

... etc.

The first line is true and each following line is just multiplied by 10 and add 3.

next consider the puzzle. Prove:

12008 = 632 x 19

120308 = 6332 x 19

1203308 = 63332 x 19

etc.

Prove it is true by recursion:

If we start with N = 120 and M = 63 we progress through the successive equations as:

(10 N + 3) x 100 + 8 = [ (10 M + 3) x 10 + 2 ] x 19

We apply recursion to the leading digits:

[ (10 N + 3) x 10 + 3 ] x 100 + 8 =  { [ (10 M + 3) x 10 + 3] x 10 + 2} x 19

10000 N + 3308 = 19000 M + 6308

10000 N = 19000 M + 3000

10 N = 19 M + 3

which, for N = 120 and M = 63 works, and so do all subsequent leading terms, as we proved from the start.

So the recursion is true.  

QED 

Edited on February 15, 2019, 2:02 am

Posted by Steven Lord on 2019-02-14 09:50:05