MathDB
Problems
Contests
National and Regional Contests
USA Contests
MAA AMC
AMC 8
1991 AMC 8
12
12
Part of
1991 AMC 8
Problems
(1)
AJHSME 1991 Problem 12
Source:
6/24/2011
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