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AMC 8
1989 AMC 8
2
2
Part of
1989 AMC 8
Problems
(1)
1989 AJHSME Problem 2
Source:
6/25/2011
2
10
+
4
100
+
6
1000
=
\frac{2}{10}+\frac{4}{100}+\frac{6}{1000} =
10
2
+
100
4
+
1000
6
=
(A)
.
012
(B)
.
0246
(C)
.
12
(D)
.
246
(E)
246
\text{(A)}\ .012 \qquad \text{(B)}\ .0246 \qquad \text{(C)}\ .12 \qquad \text{(D)}\ .246 \qquad \text{(E)}\ 246
(A)
.012
(B)
.0246
(C)
.12
(D)
.246
(E)
246
number theory
least common multiple