# General

#### RaffiO

##### New member
Write down the number that is 9 written 18 times. Divide it by 19 =
How do you work this out please?
Thanks

#### Dr.Peterson

##### Elite Member
Are you asking how to do long division, or are you expecting that there is a shortcut?

If all you need is the answer, some calculators (such as the one in Windows) can do it for you.

#### Romsek

##### Full Member
well let's see if there's anything clever that makes this trivial
$$\displaystyle n = \sum \limits_{k=0}^{17} 9\cdot 10^k = \\ \sum \limits_{k=0}^{17} (19-10) \cdot 10^k=\\ 19\sum \limits_{k=0}^{17}~10^k - 10\sum \limits_{k=0}^{17}~10^k = \\ 19 (10^{18}-1) - 10(10^{18}-1)$$

$$\displaystyle \text{Dividing the first term by 19 we obtain (10^{18}-1)}$$

$$\displaystyle \text{Is the second term even divisible by 19? We can use Fermat's Little Theorem.}\\ \text{If p is prime, for any integer a it holds that a^{p-1}-1 \equiv 0 \pmod{p}}\\~\\ \text{so 10^{18}-1 \pmod{19} = 0, so it is divisible by 19}$$

$$\displaystyle \text{I'm failing to see anything clever that will let you simply evaluate the second term divided by 19.}\\ \text{The answer isn't trivial. Maybe one of the number theory gurus will see the answer.}$$

#### Harry_the_cat

##### Senior Member
Or you could think of 9 written 18 times as $$\displaystyle 10^{18}-1$$ and then apply Fermat to show it is divisible by 19.

What is the actual question? Do you just need to show that it is divisible by 19 or do you actually want the answer to the division. If so, I think you will have to use brute force. Doesn't take that long.