# Thread: Find the last six digits of...

1. ## Find 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 $\displaystyle \frac{n}{2003}$.

2. 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....

3. Originally Posted by alexmahone
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 $\displaystyle \frac{n}{2003}$.
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.