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 $n$ .

#NumberTheory

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Comments

$3^{9} + 3^{12} + 3^{15} + 3^{n} = 3^9(1 + 3^3+3^6+3^{n-9})$ .
$(a+b)^3=a^3 + 3 a^2 b + 3 a b^2 + b^3$ .

$(3^k+1)^3 = 3^{3k} + 3^{2k+1}+3^{k+1}+1 = 1+ 3^3+3^6+3^{n-9}$
It works for $k=2, n=14$ .

Maria Kozlowska - 4 years, 7 months ago

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

Nilabha Saha - 4 years, 7 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.

Calvin Lin Staff - 4 years, 7 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.

Maria Kozlowska - 4 years, 7 months 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.

Katie Finlayson - 4 years, 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

Edwin Gray - 3 years, 7 months ago

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

Note: None of the other comments address this issue either.

Calvin Lin Staff - 3 years, 7 months ago

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

Mohammad Mainu - 4 years, 7 months ago

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

Nilabha Saha - 4 years, 7 months ago

yeap i understand

Mohammad Mainu - 4 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$