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AMC 8
1996 AMC 8
4
1996 AJHSME Problem 4
1996 AJHSME Problem 4
Source:
July 10, 2011
ratio
AMC
Problem Statement
2
+
4
+
6
+
⋯
+
34
3
+
6
+
9
+
⋯
+
51
=
\dfrac{2+4+6+\cdots + 34}{3+6+9+\cdots+51}=
3
+
6
+
9
+
⋯
+
51
2
+
4
+
6
+
⋯
+
34
=
(A)
1
3
(B)
2
3
(C)
3
2
(D)
17
3
(E)
34
3
\text{(A)}\ \dfrac{1}{3} \qquad \text{(B)}\ \dfrac{2}{3} \qquad \text{(C)}\ \dfrac{3}{2} \qquad \text{(D)}\ \dfrac{17}{3} \qquad \text{(E)}\ \dfrac{34}{3}
(A)
3
1
(B)
3
2
(C)
2
3
(D)
3
17
(E)
3
34
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