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12
AJHSME 1991 Problem 12
AJHSME 1991 Problem 12
Source:
June 24, 2011
Problem Statement
If
2
+
3
+
4
3
=
1990
+
1991
+
1992
N
\frac{2+3+4}{3}=\frac{1990+1991+1992}{N}
3
2
+
3
+
4
=
N
1990
+
1991
+
1992
, then
N
=
N=
N
=
(A)
3
(B)
6
(C)
1990
(D)
1991
(E)
1992
\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 1990 \qquad \text{(D)}\ 1991 \qquad \text{(E)}\ 1992
(A)
3
(B)
6
(C)
1990
(D)
1991
(E)
1992
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