# Thread: a rather tedious problem

1. ## 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

2. Originally Posted by andrew1990
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 = 2002×5, whose digits add up to 2. Then the number 100101001010010...10010, consisting of 10010 repeated 1001 times, will have the required property.

3. 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 $\displaystyle \frac{y}{V}$ 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 $\displaystyle \frac{2002}{14}=143$. 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 $\displaystyle \frac{16016}{2002}=8$, 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?