Determine all possible value(s) of a 15-digit base ten positive integer N which is constituted entirely by the digits 2 and 3, such that N is divisible by the base ten repunit R₇.

Consider the 15-digit number composed entirely of 2's = 2*R(15)
If you divide it by R(7) it has a remainder of 2.

We need to increase some of the digits by one (from 2 to 3) in order to increase the remainder to R(7) or a multiple of R(7).

If you increase the number by one in the 10^n place for n={0,1,2,3,4,5,6} you increase the remainder by 10^n.  For n={7,8,9,10,11,12,13} the increase is by 10^(n-7).  For n=14 (the leading digit) the increase is by by 10^0 = 1.
[i.e. it is modular]

So the remainder in each of the seven possible positions can be increased by 0,1, or 2 only (except the last could also be increased by 3) 

Clearly then, the only possible multiple of R(7) is 2.

1000000+
100000+
10000+
1000+
100+
10+
1000000+
100000+
10000+
1000+
100+
10+
2 = 2222222
So the number needs 3 in the 10^n places for n={1,2,3,4,5,6,8,9,10,11,12,13}  (all but n={0,7,14})

The number is 233333323333332

Posted by Jer on 2010-01-09 20:16:05