Let \(f_1, \dots, f_d\) be polynomials in \(\mathbb{Z} [x_1, \dots, x_d]\) with respective degrees \(k_1, \dots, k_d\), and write \[ J({\mathbf f}, {\mathbf x})= \text{det} \biggl( {{\partial f_j} \over {\partial x_i}} ({\mathbf x}) \biggr)_{1\leq i,j\leq d}. \] When \(p\) is a prime number and \(s\) is a natural number, let \({\mathcal N} ({\mathbf f}; p^s)\) denote the number of solutions of the simultaneous congruences \(f_j (x_1, \dots, x_d) \equiv 0\pmod {p^s}\), with \(1\leq x_i\leq p^s\) \((1\leq i, j\leq d)\) and \((J({\mathbf f}; {\mathbf x}),p)=1\). In this paper the author uses an elementary argument to prove that \[ {\mathcal N}({\mathbf f}; p^s)\leq k_1 \cdots k_d. \] The result is best possible, in view of the example \(f_j ({\mathbf x})= x_j^{kj} -1\) for \(1\leq j\leq d\). The author also provides a sketch of an alternative proof of the above result by using Bézout’s theorem.