If $10^{68}$ is divided by $13$, the remainder is

Let us calculate the first several powers of 10 divided by 13 to find the pattern:

Power	Division	Quotient + Remainder
$10^1$	$10 \div 13$	$0$ remainder $10$
$10^2$	$100 \div 13$	$7$ remainder $9$
$10^3$	$1000 \div 13$	$76$ remainder $12$
$10^4$	$10000 \div 13$	$769$ remainder $3$
$10^5$	$100000 \div 13$	$7692$ remainder $4$
$10^6$	$1000000 \div 13$	$76923$ remainder $1$
$10^7$	$10000000 \div 13$	$769230$ remainder $10$

The cycle repeats every 6 powers: 10, 9, 12, 3, 4, 1.

This means that we can find the general form for every $10^x$ divided by 13.

We know that $x=68$ in this case. Hence, breaking $10^{68}$ in the general form will allow us to map it to the remainder.

General Form $(k \in \text{W})$	Example	Remainder
$10^{6k+1}$	$10^1$	$10$
$10^{6k+2}$	$10^2$	$9$
$10^{6k+3}$	$10^3$	$12$
$10^{6k+4}$	$10^4$	$3$
$10^{6k+5}$	$10^5$	$4$
$10^{6k}$	$10^6$	$1$

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We need to find where $68$ fits in the cycle of $6$ (i.e. find the highest multiple of $6$ before $68$):

$\Rightarrow 68 = 6(11) + 2$

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Hence, $10^{68} = 10^{6(11) + 2} = 10^{6k + 2}$

From the table, we see the remainder at $10^{6k + 2}$ is $\boxed{9}$

It means that the remainder of $10^{68}$ is the same as $10^2$. You can use this pattern-finding approach (cyclicity) for any numbers!