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

numbers
2016-12-19 18:35:19
gajanand
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 $2$

(0) (0)
2016-12-24 13:22:45
lakhan
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 $(b)$
$13^{73}+14^3 \equiv (8+5) \equiv 13 \equiv 2 \pmod{11}$