Question

The least non-negative remainder when 3⁵¹ is divided by 7 is:

• A. 2
• B. 3
• C. 6
• D. 5

Hint:

In modular arithmetic, powers of numbers often repeat in cycles. This is called the periodicity of powers in modular arithmetic. Identifying the cycle can significantly simplify calculations, especially when dealing with large exponents. For example, instead of calculating \( 3^{51} \) directly, you can reduce the exponent modulo the length of the cycle (in this case, 6) to find an equivalent smaller exponent and simplify the calculation.

Answer 1

The Correct Option is C

To find the least non-negative remainder when \(3^{51}\) is divided by 7, we use Fermat's Little Theorem which states that if \(p\) is a prime number and \(a\) is an integer not divisible by \(p\), then \(a^{p-1} \equiv 1 \pmod{p}\). 

In this case, \(p = 7\) and \(a = 3\), so according to the theorem, \(3^{6} \equiv 1 \pmod{7}\).

This means any power of 3 that is a multiple of 6 will have a remainder of 1 when divided by 7.

First, express 51 in terms of multiples of 6:

\(51 = 6 \times 8 + 3\).

Thus, we can write:

\(3^{51} = (3^6)^8 \times 3^3\).

Since \(3^6 \equiv 1 \pmod{7}\), we have:

\((3^6)^8 \equiv 1^8 \equiv 1 \pmod{7}\).

Therefore:

\(3^{51} \equiv 1 \times 3^3 \equiv 3^3 \pmod{7}\).

Next, calculate \(3^3\):

\(3^3 = 27\).

Now, find the remainder of 27 when divided by 7:

\(27 \div 7 = 3\) with a remainder of 6.

Therefore:

\(3^3 \equiv 6 \pmod{7}\).

Thus, the least non-negative remainder when \(3^{51}\) is divided by 7 is: 6

Answer 2

Use modular arithmetic to calculate \( 3^{51} \mod 7 \).

First, observe the powers of 3 modulo 7: \[ 3^1 \equiv 3 \mod 7, \quad 3^2 \equiv 9 \equiv 2 \mod 7, \quad 3^3 \equiv 6 \mod 7, \quad 3^4 \equiv 18 \equiv 4 \mod 7. \] Further, \[ 3^5 \equiv 12 \equiv 5 \mod 7, \quad 3^6 \equiv 15 \equiv 1 \mod 7. \]

Step 1: Identify the cycle:

Notice that the powers of 3 modulo 7 repeat cyclically every 6 steps: \[ 3, 2, 6, 4, 5, 1. \] Thus, the powers of 3 modulo 7 follow a repeating cycle with a period of 6.

Step 2: Find \( 51 \mod 6 \):

Since the powers of 3 modulo 7 repeat every 6 terms, we calculate: \[ 51 \mod 6 = 3. \] This means that \( 3^{51} \equiv 3^3 \mod 7 \).

Step 3: Calculate \( 3^3 \mod 7 \):

From the earlier observation, we know: \[ 3^3 \equiv 6 \mod 7. \]

Conclusion: The least non-negative remainder when \( 3^{51} \) is divided by 7 is 6.