M is a 2003-digit positive decimal integer having the form: 12.....21
where, the first digit is 1, followed by the digit 2 repeated precisely 2001 times, and the last digit is 1.

Determine the highest power of 11 that evenly divides M.

(In reply to re: Solution by Charlie)

 10   B=1
 20   for I=1 to 2001
 30     B=B*10+2:N=10*B+1
 40     Num=N:Ct=0
 50     while (Num@11=0)
 60       Ct=Ct+1
 70       Num=Num//11
 80     wend
 90     print I;Ct
100   next

ends with:

. . .

1975  2
1976  1
1977  2
1978  1
1979  3
1980  1
1981  2
1982  1
1983  2
1984  1
1985  2
1986  1
1987  2
1988  1
1989  2
1990  1
1991  2
1992  1
1993  2
1994  1
1995  2
1996  1
1997  2
1998  1
1999  2
2000  1
2001  3

confirming the solution.

Posted by Charlie on 2009-08-01 16:30:59