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Bunuel
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As they are consequative we can write it so: 3^r(1+3^1+3^2)=3^r*13
So answer is 13 (E)
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Bunuel
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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Bunuel
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

Solution:

  • Questions like this is perfect for number picking
  • Especially because the answer options are all in whole number and none of them are none of the above or cannot be determined

  • So, let us assume \(r=0\), \(s=1\) and \(t=2\)
  • So, \(3^r + 3^s + 3^t=3^0+3^1+3^2=1+3+9=13\)
  • So, the greatest prime factor of \(3^r + 3^s + 3^t\) is 13

Hence the right answer is Option E
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