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RaffiO

New member
Joined
Jan 28, 2020
Messages
1
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
Joined
Nov 12, 2017
Messages
6,204
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
Joined
Nov 16, 2013
Messages
784
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
Joined
Mar 16, 2016
Messages
1,801
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.o_O
 
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