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}} | ||
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Latest revision as of 11:16, 13 August 2019
Problem
Solution
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
