# Math Help - Midterm review

1. ## Midterm review

Agh, I've been working on this question for an hour and can't seem to make any headway. I seem to be completely lost on it.

a.) Show that the positive integers less than 11, except 1 and 10 can be split into pairs of integers such that each pair consists of integers that are inverses of each other modulo 11.

b.) use part (a) to show that 10! = -1(mod 11)

Any help would be appreciated

2. Originally Posted by Niotsueq
Agh, I've been working on this question for an hour and can't seem to make any headway. I seem to be completely lost on it.

a.) Show that the positive integers less than 11, except 1 and 10 can be split into pairs of integers such that each pair consists of integers that are inverses of each other modulo 11.

b.) use part (a) to show that 10! = -1(mod 11)
For a), you can find out by trial and error that, for example, $2\times6 = 12\equiv1\!\!\!\pmod{11}$. Therefore 2 and 6 are inverses of each other. So also are 7 and 8, because $7\times8 = 56\equiv1\!\!\!\pmod{11}$.

For b), $10! = 1\times 2\times3\times\cdots\times10 = 1\times(2\times6)\times\cdots\times(7\times8)\time s\cdots\times10$ (pair the numbers off as in part a)).

3. Hello, Niotsueq!

a) Show that the positive integers less than 11, except 1 and 10,
can be split into pairs of integers such that each pair consists of integers
that are inverses of each other modulo 11.
. . . . . $\begin{array}{ccc}2\cdot6 &\equiv & 1\text{ (mod 11)} \\3\cdot4 &\equiv & 1\text{ (mod 11)} \\ 5\cdot9 &\equiv & 1\text{ (mod 11)} \\ 7\cdot8 & \equiv& 1 \text{ (mod 11)} \end{array}$

b) use part (a) to show that 10! = -1 (mod 11)

$10! \;\equiv\;1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cd ot8\cdot9\cdot10\,\text{ (mod 11)}$

. . . $\equiv \;1\cdot(2\cdot6)\cdot(3\cdot4)\cdot(5\cdot9)\cdot (7\cdot8)\cdot10\,\text{ (mod 11)}$

. . . $\equiv\;1\cdot 1\cdot 1\cdot 1\cdot1\cdot10\,\text{ (mod 11)}$

. . . $\equiv\;10\,\text{ (mod 11)}$

. . . $\equiv\;\text{-}1\,\text{ (mod 11)}$

4. Thanks! That's a lot easier than I thought. But, I'm not quite sure why, for instance:

2*6 = 1 (mod 11)

How does this say that 2 and 6 are inverses? I think I may be confused on the definition.

Thanks alot to both of you.

5. Originally Posted by Niotsueq
Thanks! That's a lot easier than I thought. But, I'm not quite sure why, for instance:

2*6 = 1 (mod 11)

How does this say that 2 and 6 are inverses? I think I may be confused on the definition.

Thanks alot to both of you.

$2\cdot 6 \equiv 1 \mod 11$

this shows that they are inverses of each other.

remember what an inverse is in regular multiplication?

$3 \cdot \frac{1}{3}=1$

do you see how the original means they are inverses now?

6. Originally Posted by GaloisTheory1
$2\cdot 6 \equiv 1 \mod 11$

this shows that they are inverses of each other.

remember what an inverse is in regular multiplication?

$3 \cdot \frac{1}{3}=1$

do you see how the original means they are inverses now?

So two numbers are inverses of one another mod(n) if their product is congruent to one mod(n)?