For a positive integer P drawn at random between 2 (base ten) and 2001 (base ten) inclusively, determine the probability that the sum of the digits in the base-P representation of 2009 (base ten) is equal to 11 (base ten).

    LASTYEAR = 2009

    FOR p = 2 TO 2001

        total = total + 1

        baseP$ = Base$(LASTYEAR, p)
        digitSum = SumDigits(baseP$)

        IF digitSum = 11 THEN
            count = count + 1
            PRINT p, baseP$
        END IF

    NEXT p

    PRINT
    PRINT count, total, count/total

    END

This code produced the following output:

4             133121
7             5600
10            2009
223           92
334           65
667           38
1000          29
1999          1A

8             2000          0.004

Therefore the probability is 0.4% (if we allow A to represent 10 for base 1999).

Posted by Jim Keneipp on 2010-03-13 12:54:28