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

Latest revision as of 23:10, 4 July 2013

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

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

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

1993 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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

Revision as of 21:31, 22 December 2012 (view source) Gina (talk \| contribs) m (→‎Problem 4) ← Older edit		Latest revision as of 23:10, 4 July 2013 (view source) Nathan wailes (talk \| contribs)
Line 14:		Line 14:

	{{AJHSME box\|year=1993\|num-b=3\|num-a=5}}		{{AJHSME box\|year=1993\|num-b=3\|num-a=5}}
		+	{{MAA Notice}}