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  1. #1
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    a rather tedious problem

    Hello everybody, help me please to solve this question...or give some ideas
    i)Determine, if exist, a positive number divisible by 2002, with sum of digits 2002
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  2. #2
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    Quote Originally Posted by andrew1990 View Post
    Hello everybody, help me please to solve this question...or give some ideas
    i)Determine, if exist, a positive number divisible by 2002, with sum of digits 2002
    We want to find a multiple of 2002 whose digits add up to 2002. Start with 10010 = 20025, whose digits add up to 2. Then the number 100101001010010...10010, consisting of 10010 repeated 1001 times, will have the required property.
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    Ooh, that's neat ... from what you posted, this is what I got.

    We're looking for a number N that is divisible by x, and whose digits sum to y.

    First, you consider all the divisors of y (by getting the prime factorization of y. Any combination of the prime factors will divide y).

    Second, since N is divisible by x, it is a multiple of x. So, we consider some multiples of x, and we check the sum of their digits. We are looking for those multiples of x whose digits sum up to a divisor of y.

    Say we find such a number U, and the sum of it's digits is V (and V is also a divisor of y). Then N is simply U written \frac{y}{V} times.


    For example, with x=y=2002

    The divisors of 2002 are:
    1,2,7,11,13,14,22,26,77,91,143,154,182,286,1001,20 02.

    We consider some multiples of 2002 and the sum of their digits:
    2002 4
    4004 8
    6006 12
    8008 16
    10010 2 *Opalg's example
    12012 6
    14014 10
    16016 14

    14 divides 2002, and \frac{2002}{14}=143. So, we just have to write 16016 143 times (16016160161601616016..........1601616016). That's another number that's divisible by 2002 and whose digits sum to 2002. Also, since \frac{16016}{2002}=8, I believe dividing this number (1601616016...) by 8 will yield a number 143 digits long with 8 for all it's digits (i.e. 888888...8888 where 8 is written 143 times).


    Some interesting things to think about are:

    1A) if there is a case in which no multiple of x has digits that sum up to a divisor of y.

    1B) If the above case exist, does it necessarily mean that there does not exist a number N that satisfies the condition? (Look at number 3 below)

    2) if x=y, will there always exist a number N?

    3) The method we used takes advantage of repeating the number, but what about cases with non-repeating numbers?
    Last edited by Bingk; Aug 10th 2009 at 09:38 PM. Reason: grammar
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