Difference between revisions of "1993 AJHSME Problems/Problem 4"

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{{MAA Notice}}

Latest revision as of 00:10, 5 July 2013

Problem

$1000\times 1993 \times 0.1993 \times 10 =$

$\text{(A)}\ 1.993\times 10^3 \qquad \text{(B)}\ 1993.1993 \qquad \text{(C)}\ (199.3)^2 \qquad \text{(D)}\ 1,993,001.993 \qquad \text{(E)}\ (1993)^2$

Solution

$1000\times10=10^4\\ 0.1993=1993\times10^{-4}\\ 1993\times1993\times10^{-4}\times10^4= \boxed{\textbf{(E)}\ (1993)^2}$

See Also

1993 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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