Question

Find the remainder when  \[6^{\underbrace{66\cdots6}_{100 \text{ times}}}\] is divided by 10.
 

• A. 6
• B. 2
• C. 4
• D. 8

Hint:

Units digit cycles: $2,3,7,8$ have 4-cycles; $4,9$ have 2-cycles; $5$ and $6$ are \emph{fixed} (always 5 and 6).

Answer 1

The Correct Option is A

For any positive integer exponent $k\ge1$, the last digit of $6^k$ is always $6$. Therefore $6^{\text{(any positive integer)}}\equiv 6\pmod{10}$, regardless of how large the exponent is (here it's the 100-digit number consisting only of sixes). Hence the remainder upon division by $10$ is $\boxed{6}$.

Answer 2

The last digit of a power depends only on the cyclic pattern of the base's own last digit. For a base ending in \( 6 \), examine the cycle of last digits of \( 6^1,6^2,6^3,\ldots \) directly.

1. Option A (6): Computing successive powers, \( 6^1=6,\ 6^2=36,\ 6^3=216,\ 6^4=1296,\ldots \) every one of these ends in \( 6 \). Since the cycle length is just \( 1 \), the last digit is \( 6 \) for every positive integer exponent, including the given \( 100 \)-digit exponent. This matches.
2. Option B (2): A last digit of \( 2 \) would require a base whose last-digit cycle includes \( 2 \), such as one ending in \( 2 \) or \( 8 \); this never applies to a base ending in \( 6 \), whose cycle is constant; rejected.
3. Option C (4): Similarly, \( 4 \) never appears in the cycle of last digits of powers of a number ending in \( 6 \); rejected.
4. Option D (8): The digit \( 8 \) likewise never occurs in this constant cycle of \( 6 \)'s; rejected.

Since every power of a number ending in \( 6 \) also ends in \( 6 \), the size of the exponent, however large, does not change the outcome.

Hence, the correct answer is 6.