Question

Find the remainder when 3^1993 + 11 is divided by 9.

Answer 1

\[3^{1993}+11 \] is divided by 9.

Answer 2

We know that \[a \equiv b \pmod{m}\]
means that the remainder when a is divided by m is b.
We have
\[3^{1993} \equiv 3 \pmod{9}\\
11 \equiv 2 \pmod{9}\]
Therefore, \[3^{1993}+11 \equiv 5 \pmod{9}\]
and the remainder is 5.

Answer 3

That first one should be 0, I mean. =))

Answer 4

Can it be solved without using mod? We didn't do it.

Answer 5

Well, we can do it like this:
By the division algorithm, we have
\[3^{1993}+11=9k+r \qquad 0\leq r < 9\]
where r is the remainder.
We can find r by doing the ff:
\[3^{1993}+11=9(3^{1991})+9+2=9(3^{1991}+1)+2\]
Comparing this with the above, it can be seen that the remainder is 2.

Answer 6

But didn't you calculate 5 with the previous method?

Answer 7

The first line was wrong. It was supposed to be \[3^{1993}\equiv 0 \pmod{9}\]
and adding it to the second line there will give us:
\[3^{1993}+11 \equiv 2 \pmod{9}\]

Answer 8

Okay. Thank you so much!

Answer 9

You're welcome. :D