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

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<math>3^3+3^3+3^3 = 3 \times 3^3 = \boxed{\text{(A)}\ 3^4}</math>
 
<math>3^3+3^3+3^3 = 3 \times 3^3 = \boxed{\text{(A)}\ 3^4}</math>
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==See Also==
 
==See Also==
  
 
{{AJHSME box|year=1993|num-b=6|num-a=8}}
 
{{AJHSME box|year=1993|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:16, 13 August 2019

Problem

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

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

Solution

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

See Also

1993 AJHSME (ProblemsAnswer KeyResources)
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

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