If 10⁶⁸ is divided by 13, the remainder is:

The correct answer is 9.

To find the remainder when ${10}^{68}$ is divided by 13, we can use modular arithmetic and Fermat's Little Theorem.

According to Fermat's Little Theorem, if $p$ is a prime number and $a$ is an integer not divisible by $p$, then:

$a^{p-1}\equiv 1\mathrm{mod}p$

Here, the divisor $p=13$ (which is a prime number) and the base $a=10$ (which is not divisible by 13). Applying the theorem, we get:

${10}^{13-1}\equiv 1\mathrm{mod}13$

Simplifying the exponent, we have:

${10}^{12}\equiv 1\mathrm{mod}13$

Next, we express the exponent 68 in terms of 12. Dividing 68 by 12 gives a quotient of 5 and a remainder of 8:

$68=12\times 5+8$

Using this relation, we can rewrite the term ${10}^{68}$ as follows:

${10}^{68}={\left({10}^{12}\right)}^5\times {10}^8$

Now, taking the modulo 13 on both sides, we substitute ${10}^{12}\equiv 1\mathrm{mod}13$:

${10}^{68}\equiv {\left(1\right)}^5\times {10}^8\mathrm{mod}13$

This simplifies to:

${10}^{68}\equiv {10}^8\mathrm{mod}13$

To compute ${10}^8\mathrm{mod}13$, we can use the fact that $10\equiv -3\mathrm{mod}13$:

${10}^8\equiv {\left(-3\right)}^8\equiv 3^8\mathrm{mod}13$

Let us find $3^8\mathrm{mod}13$ by calculating smaller powers of 3 modulo 13:

$3^3=27$

Since $27=13\times 2+1$, we have:

$3^3\equiv 1\mathrm{mod}13$

Using this simplification, we can express $3^8$ as:

$3^8={\left(3^3\right)}^2\times 3^2$

Substituting $3^3\equiv 1\mathrm{mod}13$ gives:

$3^8\equiv {\left(1\right)}^2\times 3^2\mathrm{mod}13$

Which simplifies to:

$3^8\equiv 9\mathrm{mod}13$

Thus, the remainder when ${10}^{68}$ is divided by 13 is 9.