3 to the power

Algebra Level pending

If 3 301 + 3 300 + 3 299 = 117 3 k 3^{301}+3^{300}+3^{299} = 117\cdot3^k , what is k ? k?


The answer is 297.

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2 solutions

Mahdi Raza
Jul 21, 2020
  • Given that:

3 301 + 3 300 + 3 299 = 117 3 k 3^{301} + 3^{300} + 3^{299} = 117 \cdot 3^{\orange{k}}

  • Now we take out a common power and evaluate

3 301 + 3 300 + 3 299 = 3 299 ( 3 0 + 3 1 + 3 2 ) = 3 299 ( 13 ) = 3 297 ( 3 2 13 ) = 3 297 ( 117 ) \begin{aligned} 3^{301} + 3^{300} + 3^{299} &= 3^{299}(3^{0} + 3^{1} + 3^{2}) \\ &= 3^{299}(13) \\ &= 3^{297} \cdot (3^2 \cdot 13) \\ &= 3^{\orange{297}} \cdot (117) \end{aligned}

  • Hence when we compare

117 3 k = 117 3 297 117 \cdot 3^{\orange{k}} = 117 \cdot 3^{\orange{297}} k = 297 \implies \boxed{\orange{k = 297}}

J L
Jul 21, 2020

3 301 + 3 300 + 3 299 = 3 299 ( 3 2 + 3 1 + 3 0 ) = 3 299 13 3^{301}+3^{300}+3^{299}=3^{299}\cdot(3^2+3^1+3^0)=3^{299}\cdot13 .

Notice that 117 = 13 3 2 117 = 13\cdot3^2 , so 3 299 13 = 117 3 297 = 117 3 k 3^{299}\cdot13=117\cdot3^{297}=117\cdot3^k . Therefore, k = 297 k = 297 .

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