Ramanujan would like it

Number Theory Level 4

$\large\color{#BA33D6}3^{ \color{#20A900}n}+ \color{#BA33D6}3^{\color{#20A900}n-\color{#624F41}1}+ \color{#BA33D6}3^{\color{#20A900}n-\color{#624F41}2}+ \color{#BA33D6}3^{\color{#20A900}n-\color{#624F41}3}+\color{#BA33D6}3^{\color{#20A900}n-\color{#624F41}4}$

For a certain $n$ the expression above has $5178$ positive divisors (inclusive of $1$ and itself), compute $n$ .

The answer is 1729.

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Brian Charlesworth
Jan 16, 2015

The expression can be written as

$N = 3^{n-4}*(3^{4} + 3^{3} + 3^{2} + 3^{1} + 3^{0}) = 3^{n-4} * 121 = 3^{n-4} * 11^{2}$ .

So as both $3$ and $11$ are prime, we can conclude that $N$ has

$[(n - 4) + 1]*(2 + 1) = 3*(n - 3)$ divisors. Thus

$3*(n - 3) = 5178 \Longrightarrow n - 3 = 1726 \Longrightarrow n = \boxed{1729}$ .

Exactly how it's done! Nice solution Brian!!

Jordi Bosch - 6 years, 4 months ago

That title might be too much of a hint though :P

Daniel Liu - 6 years, 4 months ago

I disagree. I always associate Ramanujan with hideous complexity in math problems, and, to me, a mathematician's name is not a hint to solving anything.

Whitney Clark - 6 years, 4 months ago

@Whitney Clark – This might seem a bit absurd but i feel that it was very obvious as to what the answer would be because our is normal hymnal tendency to see a pattern where there isn't any

Anirudh Chandramouli - 6 years, 3 months ago

Hahaha it's true, but I didn't find anything better:P

Jordi Bosch - 6 years, 4 months ago

Thanks. I always like to see $1729$ emerge as an answer to a question. :)

Brian Charlesworth - 6 years, 4 months ago

I thought about 1729 just because of the headline haha, but I did the same way you did, Brian. Nice solution.

Matheus Abrão Abdala - 6 years, 4 months ago

Hi,

I did everything just like you and reached only: (n - 3) = 5178. Why is there the extra 3? Thanks in advance!

Farouk Yasser - 6 years, 4 months ago

When the prime factorization of a positive integer $N$ is

$p_{1}^{a_{1}} * p_{2}^{a_{2}} * p_{3}^{a_{3}} * ..... * p_{k}^{a_{k}}$

then the number of divisors of $N$ will be

$(a_{1} + 1)*(a_{2} + 1)*(a_{3} + 1)*........*(a_{k} + 1)$ .

So in this case, as $N = 3^{n-4}*11^{2}$ the number of divisors will be

$((n - 4) + 1)*(2 + 1) = (n - 3)*3$ ,

as found in my solution. Hope that helps. :)

Brian Charlesworth - 6 years, 4 months ago

Ahaaaa... Thanks a lot for helping :))

Farouk Yasser - 6 years, 4 months ago

Because of $11^2$ .

Roman Frago - 6 years, 4 months ago

William Isoroku
Jan 17, 2015

Let $3^n=x$ so the equation becomes:

$x+\frac{x}{3}+\frac{x}{9}+\frac{x}{27}+\frac{x}{81}$

Which is $\frac{121x}{81}\longrightarrow 11^2(3^{n-4})$

Since $11$ and $3$ are both prime, we can now find the number of positive divisors.

Using Euler's method of finding the number of positive divisors, add 1 to each exponent in the prime factored form and multiply them:

$(2+1)(n-4+1)=5178\longrightarrow n=\boxed{1729}$

did it the same way

Mardokay Mosazghi - 6 years, 4 months ago

Saraswathy Rm
Jan 22, 2015

num = 3^(n-4) + 3^(n-3) + 3^(n-2) + 3^(n-1) + 3^n

num = 3^n-4 ( 1+3+9+27+81)

num = 3^(n-4 )(121)

It follows from prime factorisation that :

the factors of 121 are 11,121 (i.e 2 factors, excluding 1)

and that of 3^(n-4) are 3,3^2...........3^n-4 (i.e n-4 factors, excluding 1)

Factor combinations are made by taking one factor from each of the above set

combinations = 2(n-4) = 2n-8

taking the factors themselves from the 2 set and the #1 , we get

n-4 + 2 + 1

Total number of factors = 2n-8+n-4+2+1 = 3n-9

5178 = 3n-9

implies n = 1729

PS - 1729 is the Ramanujam number

1729 = 10^3 + 9^3

1729 = 12^3 + 1^3

Ramanujan would like it

The answer is 1729.

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