## 13 -11

- samiul_samin
**Posts:**999**Joined:**Sat Dec 09, 2017 1:32 pm

### 13 -11

$N$ is a positive integer. If $11N$ divides ($13^N-1$)then, what is the minimum

value of $N$?

value of $N$?

- samiul_samin
**Posts:**999**Joined:**Sat Dec 09, 2017 1:32 pm

### Re: 13 -11

Anyone please give any hint at least.

- ahmedittihad
**Posts:**181**Joined:**Mon Mar 28, 2016 6:21 pm

### Re: 13 -11

1. You won't get a hint just after $1$ minute of posting that problem.

2. Here's the hint, Use fermat's little theorem.

2. Here's the hint, Use fermat's little theorem.

Frankly, my dear, I don't give a damn.

### Re: 13 -11

Easy to prove by

$11|13^N-1$ We write:

$13^N-1 \equiv 0 \Rightarrow 13^N \equiv 1 \Rightarrow (11+2)^N \equiv 1$

We can write the expression $(11+2)^N=11^N+{{N} \choose {1}} 11^{N-1}.2+{{N} \choose {2}} 11^{N-2}.2^2+...+{{N} \choose {N-1}} 11.2^{N-1}+2^N$

Here, all terms are divisible by $11$ except $2^N$

So, we can write: $2^N \equiv 1$

Applying

$2^{10} \equiv 1$

Applying

$2^{GCD(N,10)} \equiv 1$

Assume that,

Here,$2^2-1$, $2^5-1$ aren't divisible by $11$

So, $2^10 \equiv 1$

Then the minimum value of $N=10$

**Modular Arithmetic**$11|13^N-1$ We write:

$13^N-1 \equiv 0 \Rightarrow 13^N \equiv 1 \Rightarrow (11+2)^N \equiv 1$

**mod**$(11)$We can write the expression $(11+2)^N=11^N+{{N} \choose {1}} 11^{N-1}.2+{{N} \choose {2}} 11^{N-2}.2^2+...+{{N} \choose {N-1}} 11.2^{N-1}+2^N$

Here, all terms are divisible by $11$ except $2^N$

So, we can write: $2^N \equiv 1$

**mod**$(11)$.....**(1)**Applying

**Fermat's Little Theorem**we get:$2^{10} \equiv 1$

**mod**$(11)$......**(2)**Applying

**GCD Lemma**for**(1)**and**(2)**, we get:$2^{GCD(N,10)} \equiv 1$

**mod**$(11)$Assume that,

**GCD**$(N,10)=d$ Then, $d$ divides $10$. So, $d=2,5,10$Here,$2^2-1$, $2^5-1$ aren't divisible by $11$

So, $2^10 \equiv 1$

**mod**$(11)$Then the minimum value of $N=10$

**I think it is the correct answer**- samiul_samin
**Posts:**999**Joined:**Sat Dec 09, 2017 1:32 pm

### Re: 13 -11

How can I getTasnood wrote: ↑Sun Feb 18, 2018 10:08 amEasy to prove byModular Arithmetic

$11|13^N-1$ We write:

$13^N-1 \equiv 0 \Rightarrow 13^N \equiv 1 \Rightarrow (11+2)^N \equiv 1$mod$(11)$

We can write the expression $(11+2)^N=11^N+{{N} \choose {1}} 11^{N-1}.2+{{N} \choose {2}} 11^{N-2}.2^2+...+{{N} \choose {N-1}} 11.2^{N-1}+2^N$

Here, all terms are divisible by $11$ except $2^N$

So, we can write: $2^N \equiv 1$mod$(11)$.....(1)

ApplyingFermat's Little Theoremwe get:

$2^{10} \equiv 1$mod$(11)$......(2)

ApplyingGCD Lemmafor(1)and(2), we get:

$2^{GCD(N,10)} \equiv 1$mod$(11)$

Assume that,GCD$(N,10)=d$ Then, $d$ divides $10$. So, $d=2,5,10$

Here,$2^2-1$, $2^5-1$ aren't divisible by $11$

So, $2^10 \equiv 1$mod$(11)$

Then the minimum value of $N=10$

I think it is the correct answer

**mod**$11N$?

### Re: 13 -11

**I missed the middle stamp!**

However a simple change will make the solution correct.

Last edited by Tasnood on Sun Feb 18, 2018 11:05 am, edited 1 time in total.

- samiul_samin
**Posts:**999**Joined:**Sat Dec 09, 2017 1:32 pm