Bunuel wrote:
If r, s and t are consecutive integers, what is the greatest prime factor of 3^r + 3^s + 3^t ?
A) 3
B) 5
C) 7
D) 11
E) 13
Let’s say r = 1, s = 2, and t = 3; then we have:
3^1 + 3^2 + 3^3 = 3^1(1 + 3 + 9) = 3(13)
Let’s say r = 2, s = 3, and t = 4; then we have:
3^2 + 3^3 + 3^4 = 3^2(1 + 3 + 9) = 3^2(13)
We see that regardless of the 3 consecutive positive integers we select, the largest prime will always be 13 since the other factor is always a power of 3.
Answer: E
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66% (01:10) correct 
