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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

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  • $\begingroup$ What are your thoughts about this question? Can you add your attempt in the main question body? $\endgroup$ Nov 23, 2017 at 16:52
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    $\begingroup$ Use Fermat's little theorem $\endgroup$
    – gambler101
    Nov 23, 2017 at 16:56
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    $\begingroup$ Or, just do the indicated arithmetic: the answer is 4, $\endgroup$
    – user247327
    Nov 23, 2017 at 16:58
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    $\begingroup$ Hint : Use $3^3\equiv 1\mod 13$ and $4^3\equiv -1\mod 13$. The result is $0$. $\endgroup$
    – Peter
    Nov 23, 2017 at 18:10
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    $\begingroup$ @JaideepKhare No one in particular. Or to myself? $\endgroup$ Nov 24, 2017 at 10:09

2 Answers 2

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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 $0$

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$13$ 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 $13$.

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