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AMC 8
1999 AMC 8
24
1999 AMC 8 #24
1999 AMC 8 #24
Source:
June 17, 2011
Problem Statement
When
1999
2
0
0
0
1999{}^2{}^0{}^0{}^0
1999
2
0
0
0
is divided by
5
5
5
, the remainder is
(A)
4
(B)
1
(C)
2
(D)
3
(E)
0
\text{(A)}\ 4\qquad\text{(B)}\ 1\qquad\text{(C)}\ 2\qquad\text{(D)}\ 3\qquad\text{(E)}\ 0
(A)
4
(B)
1
(C)
2
(D)
3
(E)
0
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