# 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,

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

so that

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

But,

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

So,

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

But,

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

So, we may finally conclude that

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

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.

$1111 = 123\times9+4$

$\Rightarrow 1111 \equiv 4 \mod 9$

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

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

$\equiv 7\mod 9$

Hello kturfHere's an alternative approach.

$1111 = 123\times9+4$

$\Rightarrow 1111 \equiv 4 \mod 9$

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

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

$\equiv 7\mod 9$
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 $n$ between 0 and 8 such that: . $1111^{2222} \:\equiv\:n\text{ (mod 9)}$

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

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

We find that:

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

The remainders move through a 3-step cycle.

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

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