# Thread: Modular Arithmetic with Exponent

1. ## Modular Arithmetic with Exponent

Determine n between 0 and 8 such that 1111^2222 equivalent to n mod 9.

I need to show the steps.

Thanks!

2. Originally Posted by kturf
Determine n between 0 and 8 such that 1111^2222 equivalent to n mod 9.

I need to show the steps.

Thanks!
Merely note that,

$\displaystyle 2222=2220+2=370\cdot6+2=370\phi\left(9\right)+2$

so that

$\displaystyle 1111^{2222}\equiv 1111^2\text{ mod }9$.

But,

$\displaystyle 1111=11\cdot101\equiv 2\cdot101\text{ mod }9$
.

So,

$\displaystyle 1111^{2222}\equiv 202^2\text{ mod }9$.

But,

$\displaystyle 202^2=\left(198+4\right)^2=198^2+2\cdot4\cdot198+4 ^2$. But, $\displaystyle 198\equiv 0\text{ mod }9$
.

So, we may finally conclude that

$\displaystyle 1111^{2222}\equiv 198^2+2\cdot4\cdot198+4^2\equiv 4^2\equiv 7\text{ mod }9$

3. Hello kturf
Originally Posted by kturf
Determine n between 0 and 8 such that 1111^2222 equivalent to n mod 9.

I need to show the steps.

Thanks!
Here's an alternative approach.

$\displaystyle 1111 = 123\times9+4$

$\displaystyle \Rightarrow 1111 \equiv 4 \mod 9$

$\displaystyle \Rightarrow 1111^2 \equiv 4^2 \mod 9$
$\displaystyle \equiv 7 \mod 9$
$\displaystyle \Rightarrow 1111^3 \equiv 4\times7 \mod 9$
$\displaystyle \equiv 1 \mod 9$
And $\displaystyle 2222 = 740\times 3 +2$

$\displaystyle \Rightarrow 1111^{2222} \equiv \Big(1111^{3}\Big)^{740} \times 1111^2 \mod 9$
$\displaystyle \equiv 1 \times 7 \mod 9$

$\displaystyle \equiv 7\mod 9$

Hello kturfHere's an alternative approach.

$\displaystyle 1111 = 123\times9+4$

$\displaystyle \Rightarrow 1111 \equiv 4 \mod 9$

$\displaystyle \Rightarrow 1111^2 \equiv 4^2 \mod 9$
$\displaystyle \equiv 7 \mod 9$
$\displaystyle \Rightarrow 1111^3 \equiv 4\times7 \mod 9$
$\displaystyle \equiv 1 \mod 9$
And $\displaystyle 2222 = 740\times 3 +2$

$\displaystyle \Rightarrow 1111^{2222} \equiv \Big(1111^{3}\Big)^{740} \times 1111^2 \mod 9$
$\displaystyle \equiv 1 \times 7 \mod 9$

$\displaystyle \equiv 7\mod 9$
One of us started with the base and the other with the exponent! Haha! You solution is probably better though since you didn't need to invoke Euler.

5. Hello, kturf!

Determine $\displaystyle n$ between 0 and 8 such that: .$\displaystyle 1111^{2222} \:\equiv\:n\text{ (mod 9)}$

Since $\displaystyle 1111 \,\equiv\,4\text{ (mod 9)}$, we have: .$\displaystyle 4^{2222}\:\equiv\:n\text{ (mod 9)}$

We want the remainder when $\displaystyle 4^{2222}$ is divided by 9.

We find that:

. . $\displaystyle \begin{array}{ccc}4^1 & \equiv & 4\text{ (mod 9)} \\ 4^2 &\equiv& 7\text{ (mod 9)} \\ 4^3 &\equiv& 1\text{ (mod 9)} \\ 4^4 &\equiv& 4\text{ (mod 9)} \\ \vdots && \vdots \end{array}$

The remainders move through a 3-step cycle.

Since $\displaystyle 2222\,\equiv\,2\text{ (mod 3)}$, we have: .$\displaystyle 4^2 \,\equiv\,7\text{ (mod 9)}$

Therefore: .$\displaystyle 1111^{2222} \:\equiv\:7\text{ (mod 9)}$