1993 AJHSME Problems/Problem 7: Difference between revisions

Revision as of 21:32, 22 December 2012

$3^3+3^3+3^3 =$

$\text{(A)}\ 3^4 \qquad \text{(B)}\ 9^3 \qquad \text{(C)}\ 3^9 \qquad \text{(D)}\ 27^3 \qquad \text{(E)}\ 3^{27}$

$3^3+3^3+3^3 = 3 \times 3^3 = \boxed{\text{(A)}\ 3^4}$

1993 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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All AJHSME/AMC 8 Problems and Solutions

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	== Problem 7 ==		== Problem ==

	<math>3^3+3^3+3^3 = </math>		<math>3^3+3^3+3^3 = </math>