Difference between revisions of "1993 AJHSME Problems/Problem 7"

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
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 21:30, 22 December 2012 (view source) Gina (talk \| contribs) ← Older edit		Revision as of 21:32, 22 December 2012 (view source) Gina (talk \| contribs) m (→‎Problem 7) Newer edit →
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−	== Problem 7 ==	+	== Problem ==

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