Let $\displaystyle n$ be a positive integer with all digits equal to $\displaystyle 5$ such that $\displaystyle n$ is divisible by $\displaystyle 2003$. Find the last six digits of $\displaystyle \frac{n}{2003}$.

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- Oct 16th 2008, 12:24 PMalexmahoneFind the last six digits of...
Let $\displaystyle n$ be a positive integer with all digits equal to $\displaystyle 5$ such that $\displaystyle n$ is divisible by $\displaystyle 2003$. Find the last six digits of $\displaystyle \frac{n}{2003}$.

- Oct 17th 2008, 02:00 PMAlexgadgetman
55555555555555

And I dont even know how you could use a formula... ah well.. trial and error does work sometimes.

Last six digits are 173518 as total resulting number is

27736173518

Bearing in mind you needed 6 digits at least, so number would need 9(?) 5's at least.

I just used trial and error.... (Worried) - Oct 17th 2008, 06:13 PMTing
Hello

Here is one way, I think.

If you are familiar with Napier's Bones method of multiplication then you can work 'backwards' to find the digits.

The last digit of $\displaystyle \frac{n}{2003}$ must be 5, since that is the only digit, (0 to 9), that will give you a number ending in 5 when multiplied by 3. The next digit must be 8, the only digit that will give you a number ending in 4, when multiplied by 3.

Please see the attachment below.

Can you see how to get the rest of the digits you need?

Hope that helps.