Math-o-Fun

Let 3 = x

Part 1 When ${ 3 }^{ 15 }$ is divided by 2

2 = (x-1)

$\therefore \quad When\quad { 3 }^{ 15 }\quad is\quad divided\quad by\quad 2,\quad we\quad get\quad { x }^{ 15 }\div (x-1)$

By using remainder theorem, we get the answer as 1.

Part 2 When ${ 3 }^{ 15 }$ is divided by 4

4 = (x+1)

$\therefore \quad When\quad { 3 }^{ 15 }\quad is\quad divided\quad by\quad 4,\quad we\quad get\quad { x }^{ 15 }\div (x+1)$

By using remainder theorem, we get the answer as -1. But the remainder should be positive and less than 4.

Hence adding 4 to -1 we get the answer as 3

$\boxed { \therefore \quad Final\quad answer\quad =\quad 1+3=4 }$

$\large{3 \equiv 1 \pmod{2} \\ \Rightarrow 3^{15} \equiv 1^{15} \equiv 1 \pmod{2} \\ 3 \equiv -1 \pmod{4} \\ \Rightarrow 3^{15} \equiv (-1)^{15} \equiv -1 \equiv 3 \pmod{4}}$

Hence , sum of remainders $=1+3=\boxed{4}$

Let's first of all we find the remainder when $3^{15}$ is divided by 2.

We know that $3^{15}$ is an odd number. So the remainder is $1$ .

Now we have to find the remainder when $3^{15}$ is divided by 3.

Let's observe a pattern:- $3 = 0\times4+3$ $3^{2} = 2\times4 +1$ $3^{3} = 6\times4+3$ $3^{4} = 20\times4+1$ On observing this we can conclude that:-

$3^{n} = p\times4 +1$ , if n is even.

$3^{n} = q\times4+3$ , if n is odd.( Here, $p$ and $q$ , are integers).

But here $n=15$ so remainder is 3.

So, sum of remainders is, $1+3=\boxed{4}$ .

Now I'm going to post the proof of conclusion made on observation.

Claim:- $3^{n}$ leaves remainder $3$ when divided by $4$ when $n$ is odd.

Proof:- Every odd number can be written as $2m+1$ .

Hence, $n = 2m+1$ , where $m$ is an integer. Now:- $3^{n} = 3^{2m+1}$ $= 3.3^{2m}$ $= 3.9^{m}$ $= 3.(1+8)^{m}$ $3(1+\binom{m}{1}8 + \binom{m}{2}8^{2} + \ldots )$ $3 + 3.4(\binom{m}{1}2+\binom{m}{2}16+ \ldots )$ Now this can be written as $3^{n} = 4p+3$ Similarly, we can prove that $3^{n}$ leaves remainder $1$ when divided by $4$ if $n$ is even.