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gajanand

asked 2016-12-19 18:35:19

when $13^{73}+14^{3}$ is divided by $11$ then remainder is

lakhan

answered 2016-12-24 13:22:45

We know that

Suppose $a$ is an integer and $p$ is prime. Then, according to Fermat's little theorem,

remainder$\left(a^{p-1}/p\right)=1$

Therefore,

remainder$\left(13^{11-1}/11\right)=1$

It means

remainder$\left[\left(13^{10}\right)^7/11\right]=1$

$\Rightarrow$ remainder$\left(13^{70}/11\right)=1$

So we have to solve now

$13^3/11+14^3/11$

$13^3=2197\\

14^3=2744$

We can write here

$(2197+2744)/11\\

4941/11$

Finally remainder we got

Remainder $=2$

Answer is $2$

Suppose $a$ is an integer and $p$ is prime. Then, according to Fermat's little theorem,

remainder$\left(a^{p-1}/p\right)=1$

Therefore,

remainder$\left(13^{11-1}/11\right)=1$

It means

remainder$\left[\left(13^{10}\right)^7/11\right]=1$

$\Rightarrow$ remainder$\left(13^{70}/11\right)=1$

So we have to solve now

$13^3/11+14^3/11$

$13^3=2197\\

14^3=2744$

We can write here

$(2197+2744)/11\\

4941/11$

Finally remainder we got

Remainder $=2$

Answer is $2$

Jay Patel

answered 2016-12-19 19:18:52

Ans: $2$

$13^{73}\equiv (11+2)^{73}\equiv 2^{73} \\ \equiv \left(2^5\right)^{14}×2^3 \equiv (32)^{14}×2^3 \\

\equiv (-1)^{14}×8 \equiv 8 \pmod{11} ~~\cdots(a)$

$14^{3}\equiv (11+3)^{3}\equiv 3^{3} \\ \equiv 5 \pmod{11} ~~\cdots(b)$

From $(a)$ and $(b)$

$13^{73}+14^3 \equiv (8+5) \equiv 13 \equiv 2 \pmod{11}$

Note: click here to understand about congruence relation.

$13^{73}\equiv (11+2)^{73}\equiv 2^{73} \\ \equiv \left(2^5\right)^{14}×2^3 \equiv (32)^{14}×2^3 \\

\equiv (-1)^{14}×8 \equiv 8 \pmod{11} ~~\cdots(a)$

$14^{3}\equiv (11+3)^{3}\equiv 3^{3} \\ \equiv 5 \pmod{11} ~~\cdots(b)$

From $(a)$ and $(b)$

$13^{73}+14^3 \equiv (8+5) \equiv 13 \equiv 2 \pmod{11}$

Note: click here to understand about congruence relation.

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