Question

Find the reminder when $3^{101}$ is divided by 7?

• A. 2
• B. 3
• C. 5
• D. 6

Answer 1

Answer:

C. 5

Step 1: Find the repeating pattern of remainders

We divide consecutive powers of 3 by 7 to see how the remainders behave:

• $3^1 = 3 \div 7 \implies \text{Remainder} = \mathbf{3}$

• $3^2 = 9 \div 7 \implies \text{Remainder} = \mathbf{2}$

• $3^3 = 27 \div 7 \implies \text{Remainder} = \mathbf{6}$

• $3^4 = 81 \div 7 \implies \text{Remainder} = \mathbf{4}$

• $3^5 = 243 \div 7 \implies \text{Remainder} = \mathbf{5}$

• $3^6 = 729 \div 7 \implies \text{Remainder} = \mathbf{1}$

If we continue to $3^7$ ($2187 \div 7$), the remainder goes back to 3, meaning the cycle starts over.

Step 2: Identify the Cyclicity Length

The remainders repeat in a fixed block of 6 numbers: $\{3, 2, 6, 4, 5, 1\}$.
Therefore, the cyclicity length is 6.

Step 3: Divide the exponent by the cyclicity length

Our total power is 101. We divide 101 by our cycle length of 6 to find out where this power lands in the repeating pattern:

$101 \div 6 = 16 \text{ complete cycles with a remainder of } \mathbf{5}$

Step 4: Find the final remainder

The remainder of 5 tells us that $3^{101}$ will have the exact same remainder as the $5^{\text{th}}$ step of our repeating pattern.

Looking back at our pattern from Step 1:

1. $1^{\text{st}}$ remainder = 3

2. $2^{\text{nd}}$ remainder = 2

3. $3^{\text{rd}}$ remainder = 6

4. $4^{\text{th}}$ remainder = 4

5. $5^{\text{th}}$ remainder = 5

Thus, the final remainder when $3^{101}$ is divided by 7 is 5.