Super seven's

Number Theory Level 1

What is $\left({ 7 }^{ 77 }-7\right) \bmod 8$ ?

The answer is 0.

3 solutions

Jesse Nieminen
Feb 12, 2016

$7^{77} - 7 \equiv {\left(-1\right)}^{77} + 1 \equiv -1 + 1 \equiv 0$ (mod 8)

Moderator note:

Simple standard approach.

What approach is this? @BrilliantMathematics

Kaushik Chandra - 3 years, 2 months ago

Here, I am utilizing the fact that $7 \equiv -1 \pmod{8}$ which means that $7$ is equivalent to $-1$ in addition and multiplication modulo $8$ .

Jesse Nieminen - 3 years, 2 months ago

Swafim Li
Mar 14, 2018

According to binomial theorem

$7^{77}-1 \\=(8-1)^{77}-7 \\=\sum_{r=0}^{77}{\left( C_{77}^r\cdot 8^{77-r}(-1)^r\right)-7 } \\=\sum_{r=0}^{76}{\left( C_{77}^r\cdot 8^{77-r}(-1)^r\right)-1-7 } \\=\sum_{r=0}^{76}{\left( C_{77}^r\cdot 8^{77-r}(-1)^r\right)-8 }$

So $\left(7^{77}-7 \right)\bmod 8 =0$

Kartik Sharma
Aug 29, 2014

Using Euler's theorem, as 7 and 8 are co-prime,

${7}^{\phi(8)} = 1 mod 8$

${7}^{4} = 1 mod 8$

${7}^{76} = 1 mod 8$

${7}^{77} = 7 mod 8$

${7}^{77} - 7 = 7-7 mod 8$

${7}^{77} - 7 = 0 mod 8$

Anyway one can also write :

$7=-1\quad mod\quad 8$

${ 7 }^{ 77 }={ (-1) }^{ 77 }\quad mod\quad 8$

$\Rightarrow { 7 }^{ 77 }=\quad -1\quad mod\quad 8$

${ 7 }^{ 77 }-7=\quad -8\quad mod\quad 8$

${ 7 }^{ 77 }-7=\quad 0\quad mod\quad 8$

No need of euler theorom.

Ronak Agarwal - 6 years, 9 months ago

Good solution @Ronak Agarwal

shivamani patil - 6 years, 9 months ago

Done the Same

Jayakumar Krishnan - 6 years, 9 months ago

Hey, actually the solution writing guide was for Level 3 and so, we should post solutions based on the level 3. :P

Kartik Sharma - 6 years, 9 months ago

i hav a doubt,

why cant we do like

7^7^11= 7^77,

7^7 = 1(mod 8),

1^11 = 1,

So, 7^77-7 = 1-7 = (-6),

-6 = 8*0-6, hence remainder should be -6, i hav not learnt modulus, understanding my means of self solving, pls tell me where i hav gone wrong.

Mohammed Ali - 6 years, 6 months ago

dear 7^2 is =1 so in remender probes power to power must be be the multiple of 2(for this case) and 11 is not the multiple of 2 so it would be like (7^2)^10=1 and single 7 is left so it is 1*7 (mod 8)=7 hence 7-7(mod 8)=0

Rajkumar Saini - 5 years, 8 months ago

Super seven's

The answer is 0.

3 solutions

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