To find the remainder of \( 10^{68} \) divided by 13, we use Fermat's Little Theorem. This theorem states that if \( p \) is a prime number and \( a \) is an integer not divisible by \( p \), then \( a^{p-1} \equiv 1 \ (\text{mod} \ p) \).
In this case:
\( a = 10 \) and \( p = 13 \). Therefore, \( 10^{12} \equiv 1 \ (\text{mod} \ 13) \).
We can rewrite \( 10^{68} \) using powers of 12:
Dividing 68 by 12 gives 5 with a remainder of 8. So, \( 68 = 5 \times 12 + 8 \).
This means \( 10^{68} = (10^{12})^5 \times 10^8 \). Using the theorem, this simplifies to \( 1^5 \times 10^8 \equiv 10^8 \ (\text{mod} \ 13) \).
Now, we calculate \( 10^8 \pmod{13} \):
\( 10^2 = 100 \equiv 9 \ (\text{mod} \ 13) \)
\( 10^4 = (10^2)^2 \equiv 9^2 = 81 \equiv 3 \ (\text{mod} \ 13) \)
\( 10^8 = (10^4)^2 \equiv 3^2 = 9 \ (\text{mod} \ 13) \)
Therefore, the remainder when \( 10^{68} \) is divided by 13 is 9.