# 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
1 year, 5 months 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 $$k=2, n=14$$.

- 1 year, 5 months ago

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.

Staff - 1 year, 5 months ago

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.

- 1 year, 5 months ago

Thank you very much. You literally releived me from an enthralling problem!

- 1 year, 5 months ago

We can find n by successive substitution as follows(without thinking too much). Let 3^9 + 3^12 + 3^15 + 3^n = t^3. t must be divisible by 3, so let t =3r, and substitute: 3^9 + 3^12 + 3^15 + 3^n = (3r)^3 = 3^3*r^3. Dividing by 3^3, 3^6 + 3^9 + 3^12 +3^(n-3) = r^3. r must be divisible by 3, so let r = 3s, and substitute. Saving the reader a bit, s must be divisible by 3, so let s = 3u. We arrive at the following equation: 1 + 3^3 + 3^6 + 3^(n-9) = u^3. u must be of the form u = 3x + 1. Cubing, u^3 = 27x^3 + 27x^2 + 9x + 1, so 1 + 3^3 + 3^6 + 3^(n-9) = 27x^3 + 27x^2 + 9x + 1. Letting x = 3, 1 + 27 +729 + 3^(n-9) = 729 + 243 +27 + 1, or 3^(n-9) = 243 , or n-9 =5, n = 14. Ed Gray

- 5 months, 3 weeks ago

How do you know that's all of the possible values? IE Why can't there be another $$x$$ value that works?

Staff - 5 months, 2 weeks ago

I'd start by thinking that that looks like a trinomial expansion. So let's take our cubed result and write it as 3^3 + a.

So we have 3^9 + 3^12 + 3^15 + 3^n = (3^3 + a)^3

=> 3^9 + 3^12 + 3^15+3^n = 3^9 + 3a3^6 + 3a^2 3^3 + a^3

=> 3^9 + 3^12 + 3^15+3^n = 3^9 + a3^7 + a^2 3^4 + a^3

Which fits if a = 3^5, and n = 14.

You can show that those are the only values by considering the graphs (of y = (fixed point + 3^x) and y=x^3) - they will only cross at one point.

- 1 year, 3 months ago

Comment deleted Jan 12, 2017

While it is true that the graphs of $$y = C + 3^x$$ will only cross the graph of $$y = x^3$$ at one point, when we require the expression to be a perfect cube, it need not just be that $$C + 3^x = x^3$$. It is actually $$C + 3^x = n^3$$ for all integers $$n$$.

Staff - 1 year, 3 months ago

Yes, I was thinking of comparing something like an exponential with a cubic progression rather than solving both for x, but you're right, that doesn't fly where there is no relationship between the two integers. Interesting.

- 1 year, 3 months ago

3^9,3^12,3^15.....9+3=12+3=15+3=18...so,I hope the answer is 18....

- 1 year, 5 months ago

The answer is not 18. It is 14. The answer can't be found by assuming a geometric progression.

- 1 year, 5 months ago

yeap i understand

- 1 year, 5 months ago