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

Revision as of 21:31, 22 December 2012

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

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

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

1993 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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All AJHSME/AMC 8 Problems and Solutions

Revision as of 21:28, 22 December 2012 (view source) Gina (talk \| contribs) ← Older edit		Revision as of 21:31, 22 December 2012 (view source) Gina (talk \| contribs) m (→‎Problem 4) Newer edit →
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−	== Problem 4 ==	+	== Problem ==

	<math>1000\times 1993 \times 0.1993 \times 10 = </math>		<math>1000\times 1993 \times 0.1993 \times 10 = </math>