Remainder when $3^{12} + 4^{21}$ is divided by 13.

Question

What is remainder when $3^{12} + 4^{21}$ Is divided by 13? I tried this question by found binomial theorem but as there are two bases it is becoming difficult to calculate . I am a school going child and do not know higher formulae. Please help me

What are your thoughts about this question? Can you add your attempt in the main question body? — Jaideep Khare, Nov 23 '17 at 16:52
Actually I tried using binomial theorem but not getting correct answer — , Nov 23 '17 at 16:53
You should add this in main body of the question, as a question without any attempt, is most probably closed (or put on hold) here to discourage people from asking their homework, without doing anything on their own. EDIT : As I said previously, the question is put on hold. But don't worry, as soon as you add your attempt, the question will be reopened. — Jaideep Khare, Nov 23 '17 at 16:56
I'll be darned! You are right. My calculator must be getting old. — user247327, Nov 23 '17 at 18:09
Hint : Use $3^3\equiv 1\mod 13$ and $4^3\equiv -1\mod 13$. The result is $0$. — Peter, Nov 23 '17 at 18:10
I would vote to close this as a duplicate of this generic version, but I have promised not to. — Jyrki Lahtonen, Nov 23 '17 at 21:06
Which standard and where did you encounter this question? If you don't know the Relevant theory it is not reasonable that you have been asked the question. — vishal mishra, Nov 24 '17 at 18:09

score 1 · Answer 1 · answered Nov 23 '17 at 20:56

Let $a,b\in \mathbb{Z}$ and $n\in \mathbb{N}$. We say that "a is congruent to b modulo m" and write "$a\equiv b(mod m) $", if $m|(a-b)$.

So from this congruences defination $3^3=27 \Rightarrow 3^3\equiv 1(mod 13)\Rightarrow 3^{3*4}\equiv 1^4(mod 13)$ As if $a\equiv b (mod m)$ and $k\in \mathbb{N}$ then $a^k\equiv b^k(mod m)$

so $3^{12}\equiv 1(mod 13)$

Again Similarly $4^3\equiv -1(mod 13)\Rightarrow 4^{21}\equiv -1(mod 13)$

Now $3^{12} +4^{21}\equiv 0(mod 13)$

13 divide $3^{12} +4^{21}$ so reminder is $0$

score 0 · Accepted Answer · answered Nov 23 '17 at 21:11

$13$ is prime, so by Fermat's Little theorem we know that $13$ divides $a^{12}-1$, where $a$ is any number not divisible by $13$.

So $3^{12}$ has a remainder of $1$ when divided by $13$.

Then $4^{21} = 2^{42}$ has the same remainder as $2^6$ since you have $2^{12}=13k+1$ for some $k$ (again from Fermat) and so $2^{42}= 2^6(2^{12})^3 = 2^6(13k+1)^3 = 13m + 2^6$ since all terms from multiplying out the cube except the " $+1^3$ " are divisible by $13$.

Since $2^6=64$ is not large you can just add the remainders, $1+64=65$ to find the sum $3^{12}+4^{21}$ is divisible by $13$.

Remainder when $3^{12} + 4^{21}$ is divided by 13.

2 Answers2