AMC 8 1991 Problem 4

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

Dataset text NuminaMath-1.5. Quoted with source.

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