Use modular arithmetic to calculate \( 3^{51} \mod 7 \). Note that:
\[ 3^1 \equiv 3 \mod 7, \quad 3^2 \equiv 9 \equiv 2 \mod 7, \quad 3^3 \equiv 6 \mod 7, \quad 3^4 \equiv 18 \equiv 4 \mod 7. \]
Observe that the powers of 3 modulo 7 repeat cyclically every 6 steps: 3, 2, 6, 4, 5, 1.
Since \( 51 \mod 6 = 3 \), the equivalent power is \( 3^3 \). From above:
\[ 3^3 \equiv 6 \mod 7. \]
Thus, the least non-negative remainder is 6.