Question

The remainder when \(6^{1029}\) is divided by 7 is:

• A. 6
• B. 1
• C. 0
• D. 3

Answer 1

The Correct Option is A

To find the remainder when \(6^{1029}\) is divided by 7, we can use Fermat's Little Theorem. Fermat's Little Theorem states that if \(p\) is a prime number and \(a\) is an integer not divisible by \(p\), then \(a^{p-1} \equiv 1 \pmod{p}\). 

Here, \(p = 7\) and \(a = 6\). According to the theorem, \(6^{7-1} = 6^6 \equiv 1 \pmod{7}\).

We need to express \(1029\) in terms of a multiple of 6 plus a remainder, since the exponent is large. We perform division: \(1029 = 6 \times 171 + 3\).

Thus, \(6^{1029} = 6^{6 \times 171 + 3} = (6^6)^{171} \times 6^3\). As per Fermat's Theorem, \((6^6)^{171} \equiv 1^{171} = 1 \pmod{7}\).

Now, we need to calculate \(6^3 \pmod{7}\):

\(6^3 = 216\). Calculate \(216 \mod 7\):

Since \(216 / 7 = 30\) with a remainder of 6, \(6^3 \equiv 6 \pmod{7}\).

Therefore, the remainder when \(6^{1029}\) is divided by 7 is 6.

Answer 2

We are tasked to compute \(6^{1029} \mod 7\). Using Fermat's Little Theorem:

Step 1: Apply Fermat's Little Theorem.

Fermat's Little Theorem states:

\[a^{p-1} \equiv 1 \pmod{p},\]

for a prime p and an integer a not divisible by p. Here \(a = 6\) and \(p = 7\). Since 6 is not divisible by 7, we have:

\[6^6 \equiv 1 \pmod{7}.\]

Step 2: Simplify the exponent.

To compute \(6^{1029} \mod 7\), divide 1029 by 6 (the exponent cycle length from Fermat's theorem):

\[1029 \div 6 = 171 \text{ remainder 3}.\]

Thus:

\[6^{1029} \equiv 6^3 \pmod{7}.\]

Step 3: Compute \(6^3 \mod 7\).

Now calculate \(6^3 \mod 7\):

\[6^3 = 216.\]

Find the remainder when 216 is divided by 7:

\[216 \div 7 = 30 \text{ remainder 6}.\]

Thus:

\[6^3 \equiv 6 \pmod{7}.\]

Final Answer:

\[6\]