1970 AHSME Problem 34

The greatest integer that will divide 13511, 13903, and 14589 and leave the same remainder is

\begin{array}{l} \textbf{(A) }28\qquad \textbf{(B) }49\qquad \textbf{(C) }98\qquad\\ \textbf{(D) }\text{an odd multiple of }7\text{ greater than }49\qquad\\ \textbf{(E) }\text{an even multiple of }7\text{ greater than }98 \end{array}

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