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4
AJHSME 1991 Problem 4
AJHSME 1991 Problem 4
Source:
June 24, 2011
Problem Statement
If
991
+
993
+
995
+
997
+
999
=
5000
−
N
991+993+995+997+999=5000-N
991
+
993
+
995
+
997
+
999
=
5000
−
N
, then
N
=
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
(A)
5
(B)
10
(C)
15
(D)
20
(E)
25
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