looks too easy right?

Number Theory Level 3

$3 + 3^{2} + 3^{3} + 3^{4} + 3^{5}+ \ldots. + 3^{104}$ is divisible by

5 and 2 only 3 and 1 only 1 , 2 , 3 and 5 (5-2) only 1 only 3 only 1 , 2 and 3 0nly

1 solution

Chew-Seong Cheong
Mar 2, 2015

$\begin{aligned} \text{Let } S & = 3+3^2+3^3+3^4+3^5+... \\ & = 3(1+3+3^2+3^3+3^4+...) \\ & = \dfrac {3(3^{104}-1)}{3-1} \\ & = \dfrac {3(3^{104}-1)}{2} \end{aligned}$

$S$ is of course divisible by $1$ .
Since $S$ is a whole number, $\Rightarrow 3^{104}-1 \equiv 0 \pmod {2}$ .
Since $3$ is a factor of $S$ , it is divisible by $3$ .
$3^{104} \equiv (3^2)^{52} \equiv (10-1)^{52} \equiv 1 \pmod {5}\Rightarrow 3^{104} - 1\equiv 0 \pmod {5}$ .

Therefore, $3+3^2+3^3+3^4+3^5+...$ is divisible by $\boxed {\text{1, 2, 3 and 5}}$ .

A brute force way to demonstrate this is simply by factorizing a bit as follows:

$3+3^2+3^3+3^4+\ldots +3^{104}\\ = 3(1+3)+3^3(1+3)+\ldots +3^{103}(1+3)\\ = 4\left(3^1+3^3+3^5+3^7+3^9+\ldots +3^{101}+3^{103}\right)\\ = 4\left(3^1(1+9)+3^5(1+9)+\ldots +3^{101}(1+9)\right)\\ = 4\times 10\times \left(3^1+3^5+\ldots +3^{101}\right)$

It is obvious that the expression inside the brackets is a whole number and we have $4$ and $10$ as factors of the original expression. This trivially implies that the expression is divisible by $2$ and $5$ .

The reason for the expression being divisible by $1$ and $3$ is the same as you provided in your solution.

Prasun Biswas - 6 years, 3 months ago

In second case , it should be proved that $S$ is divisible by $4$ because if it is only divisible by $2$ , then the $2$ in the denominator will cancel it off.

Vighnesh Raut - 6 years, 1 month ago

Good Solution sir

Sai Ram - 5 years, 11 months ago

Sir, in divisibility by $5$ , why is it congruent to $1$ $\pmod 5$ and not $-1$ $\pmod 5$

Vinayak Srivastava - 11 months, 4 weeks ago

looks too easy right?

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