How to find remainder when $ 975^{40153}$ is divided by $14$?

Question

I still find tricky this kind of problems. I tried to do solve it by factoring $14$ in $2*7$.

Then, with Fermat's Little Theorem, I find that:

$975^6\equiv 1\pmod 7$

$975^1\equiv 1\pmod 2$

How can I proceed frome here?

Answer 1

$\color\red{975^{3}}\equiv\color\red{1}\pmod{14}\implies$

$975^{40153}\equiv975^{3\cdot13384+1}\equiv(\color\red{975^{3}})^{13384}\cdot975^{1}\equiv\color\red{1}^{13384}\cdot975^{1}\equiv975\equiv9\pmod{14}$

Answer 2

$\varphi (14)=6$ and $gcd(975, 14)=1$. From Euler's theorem, we get $975^6 \equiv 1 (\mod 14)$. Now $975^{40153}\equiv 975^{6692*6+1}\equiv975\equiv9 (\mod 14)$

Answer 3

Then, $975^{40153}=975^{6692\times6+1}=(975^6)^{6922}\times975\equiv(-1)^{6922}\times975=975\equiv2\mbox{ (mod 7)}$.

Also, $975^{40153}=(975^1)^{40153}\equiv(1)^{40153}=1\mbox{ (mod 2)}$.

Using the Chinese Remainder Theorem (AKA trial and error), you would find the numbers from $0$ to $13$ which $\equiv2\mbox{ (mod 7)}$, which are $2$ and $9$, and then test for $\equiv1\mbox{ (mod 2)}$, which gives us $\underline{\underline{9}}$.

Answer 4

Using Fermat's theorem in these type of questions is not a wise decision.You may use a more generalized-Euler Theorem.

According to Euler's Theorem.

$$a^{\phi n}\equiv1\pmod n$$

Where $(\phi{n})$ is the Euler Phi function.

Now,notice that,

$$975^{\phi(14)}\equiv1\pmod{14}$$

$$\implies975^6\equiv1\pmod{14}$$

$$\implies975^{40152}\equiv1\pmod{14}\space\space\space\space\space\space\space\space\space(\text{As,40152=6}\times6692)$$

$$\implies975^{40152}\times975\equiv975\pmod{14}$$

$$\implies975^{40153}\equiv9\pmod{14}\space\space\space\space\space\space\space\space\space\space (\text{As,}975\pmod{14})=9$$

Hope this helps!!

Answer 5

$$975\equiv-5\pmod{14}\implies975^{40153}\equiv(-5)^{40153}$$

As $(-5)^3\equiv1\pmod{14}$

and as $4+0+1+5+3\equiv1\pmod3\implies40153\equiv1\pmod3$

$(-5)^{40153}\equiv(-5)^1\pmod{14}\equiv-5+14=?$