1999 AMC 8 Problem 1

(6?3) + 4 - (2 - 1) = 5 To make this statement true, the question mark between the 6 and the 3 should be replaced by

\text{(A)} \div \qquad \text{(B)}\ \times \qquad \text{(C)} + \qquad \text{(D)}\ - \qquad \text{(E)}\ \text{None of these}

Full credit goes to MAA for authoring these problems. These problems were taken on the AOPS website.