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

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Revision as of 21:52, 9 June 2011

Problem 4

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

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

Solution

$1000\times10=10^4$

$0.1993=1993\times10^{-4}$

$1993\times1993\times10^{-4}\times10^4=(1993)^2$

$\boxed{E) (1993)^2}$

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