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MAA AMC
AMC 8
1989 AMC 8
1
1
Part of
1989 AMC 8
Problems
(1)
2010 AMC 8 Problem 1-Group of 50 sums
Source:
6/25/2011
(
1
+
11
+
21
+
31
+
41
)
+
(
9
+
19
+
29
+
39
+
49
)
=
(1+11+21+31+41)+(9+19+29+39+49)=
(
1
+
11
+
21
+
31
+
41
)
+
(
9
+
19
+
29
+
39
+
49
)
=
(A)
150
(B)
199
(C)
200
(D)
249
(E)
250
\text{(A)}\ 150 \qquad \text{(B)}\ 199 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 249 \qquad \text{(E)}\ 250
(A)
150
(B)
199
(C)
200
(D)
249
(E)
250