MathDB
AJHSME 1991 Problem 4

Source:

June 24, 2011

Problem Statement

If 991+993+995+997+999=5000N991+993+995+997+999=5000-N, then N=N=
(A) 5(B) 10(C) 15(D) 20(E) 25\text{(A)}\ 5 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 15 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 25