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 $\displaystyle \frac{n}{2003}$.
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....
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.