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

Revision as of 21:28, 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
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

@@ Line 7: / Line 7: @@
 ==Solution==
-<math>1000\times10=10^4</math>
+<math>1000\times10=10^4\\
+.1993=1993\times10^{-4}\\
+\times1993\times10^{-4}\times10^4= \boxed{\textbf{(E)}\ (1993)^2}</math>
-<math>0.1993=1993\times10^{-4}</math>
+==See Also==
-<math>1993\times1993\times10^{-4}\times10^4=(1993)^2</math>
+{{AJHSME box|year=1993|num-b=3|num-a=5}}
-<math>\boxed{E) (1993)^2}</math>