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Powers Of Three, Oh Well!

Can anyone please help me solve this problem? I have been stuck with this for a long time and am unable to solve it.

The expression:

\(3^{9} + 3^{12} + 3^{15} + 3^{n} \)

is a perfect cube for some natural number \(n\). Find the value of \(n\).

Note by Nilabha Saha
3 weeks, 1 day ago

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\(3^{9} + 3^{12} + 3^{15} + 3^{n} = 3^9(1 + 3^3+3^6+3^{n-9})\).
\((a+b)^3=a^3 + 3 a^2 b + 3 a b^2 + b^3\). \((3^k+1)^3 = 3^{3k} + 3^{2k+1}+3^{k+1}+1 = 1+ 3^3+3^6+3^{n-9}\)
It works for \(k=2, n=14\). Maria Kozlowska · 3 weeks, 1 day ago

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@Maria Kozlowska Nice factorization! Are there other values of \(n\) that would work?

I can show that \( n = 3m \) would not work, but the other cases seem somewhat hairy. Calvin Lin Staff · 3 weeks ago

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@Calvin Lin I think it can be shown that the number to be cubed needs to be in a form \((9n+1)3^3\). Wolfram alpha shows just one solution. For a proper proof you might need some number theory experts on Brilliant. Maria Kozlowska · 2 weeks, 6 days ago

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@Maria Kozlowska Thank you very much. You literally releived me from an enthralling problem! Nilabha Saha · 3 weeks, 1 day ago

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3^9,3^12,3^15.....9+3=12+3=15+3=18...so,I hope the answer is 18.... Md Mainu · 2 weeks, 3 days ago

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@Md Mainu The answer is not 18. It is 14. The answer can't be found by assuming a geometric progression. Nilabha Saha · 2 weeks, 3 days ago

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@Nilabha Saha yeap i understand Md Mainu · 1 week, 4 days ago

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