When $3^{333}$ is divide by 11, the remainder is

You can find the cyclicity easily (use the calculator if available):

Power	Division	Quotient + Remainder
$3^1$	$3 \div 11$	$0$ remainder $3$
$3^2$	$9 \div 11$	$0$ remainder $9$
$3^3$	$27 \div 11$	$2$ remainder $5$
$3^4$	$81 \div 11$	$7$ remainder $4$
$3^5$	$243 \div 11$	$22$ remainder $1$
$3^6$	$729 \div 11$	$66$ remainder $3$
$3^7$	$2187 \div 11$	$198$ remainder $9$

The cycle repeats every 5 powers: 3, 9, 5, 4, 1.

This means that we can find the general form for every $3^x$ divided by 11.

We know that $x=333$ in this case. Hence, breaking $3^{333}$ in the general form will allow us to map it to the remainder.

General Form $(k \in \text{W})$	Example	Remainder
$3^{5k+1}$	$3^1$	$3$
$3^{5k+2}$	$3^2$	$9$
$3^{5k+3}$	$3^3$	$5$
$3^{5k+4}$	$3^4$	$4$
$3^{5k}$	$3^5$	$1$

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We need to find where $333$ fits in the cycle of $5$ (i.e. find the highest multiple of $5$ before $333$):

Since multiples of $5$ end in $0$ or $5$, we can break down $333 = 330 + 3$.

$\Rightarrow 333 = 5k + 3$

(where $k$ is a whole number and $5k=330$)

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Hence, $3^{333} = 3^{5k + 3}$

From the table, we see the remainder at $3^{5k + 3}$ is $\boxed{5}$

It means that the remainder of $3^{333}$ is the same as $3^3$. You can use this pattern-finding approach (cyclicity) for any numbers!