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.
To determine the least non-negative remainder of \(3^{51}\) divided by 7, we apply Fermat's Little Theorem. This theorem states that for a prime number \(p\) and an integer \(a\) not divisible by \(p\), \(a^{p-1} \equiv 1 \pmod{p}\).
With \(p = 7\) and \(a = 3\), the theorem implies \(3^{7-1} \equiv 3^6 \equiv 1 \pmod{7}\).
This congruence indicates that any power of 3 which is a multiple of 6 will yield a remainder of 1 when divided by 7.
We express the exponent 51 as a multiple of 6 plus a remainder:
\(51 = 6 \times 8 + 3\).
Consequently, \(3^{51}\) can be rewritten as:
\(3^{51} = 3^{6 \times 8 + 3} = (3^6)^8 \times 3^3\).
Given that \(3^6 \equiv 1 \pmod{7}\), we substitute this into the expression:
\((3^6)^8 \equiv 1^8 \equiv 1 \pmod{7}\).
Therefore, the congruence simplifies to:
\(3^{51} \equiv 1 \times 3^3 \equiv 3^3 \pmod{7}\).
Next, we compute \(3^3\):
\(3^3 = 27\).
Finally, we find the remainder of 27 when divided by 7:
\(27 = 7 \times 3 + 6\).
This means:
\(3^3 \equiv 6 \pmod{7}\).
Thus, the least non-negative remainder when \(3^{51}\) is divided by 7 is: 6