1
$\begingroup$

I just incurred a question which asked me to find the remainder when $41^{77}$ is divided by $7$. I just saw $41$ and then the number $6$ striked to me as $41-35=6$. I chose $35$ as it was nearest to $41$ but I think the way I solved it is wrong. For example we want to find the remainder when $35$ is divided by $6$. If someone uses this method he will end up with $1$ as he would split $35$ as $7×5$ and as $6$ is nearest to $7$ and $7-6=1$ so remainer is $1$ bur actually remainder is $5$. Please help me to get an appropriate method to solve such questions. Also such questions can be framed in a more complex way like "find the remainder obtained when $41^{77}$ is divided by $7^{21}$. So what would one do in that case? Any help is appreciated. Thank you very much.

$\endgroup$

1 Answer 1

2
$\begingroup$

You have

$$41\equiv -1 \mod 7$$

$$41^{77} \equiv (-1)^{77} \mod 7$$

$$41^{77}\equiv -1 \mod 7$$

the remainder is $6$.

$\endgroup$
1
  • $\begingroup$ bur can u also list out a method for the next part of the question $\endgroup$ Dec 15, 2016 at 18:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .