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Powers of 3

Source: 1993 AMC 8 Problem 7

June 28, 2011
AMC

Problem Statement

33+33+33=3^3+3^3+3^3 =
(A) 34(B) 93(C) 39(D) 273(E) 327\text{(A)}\ 3^4 \qquad \text{(B)}\ 9^3 \qquad \text{(C)}\ 3^9 \qquad \text{(D)}\ 27^3 \qquad \text{(E)}\ 3^{27}