MHB Prove a number is divisible by a number?

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The discussion centers on proving that 6 divides the expression 7n - 1 through mathematical induction. The initial proof attempts to establish the base case and the inductive step, but some errors in logic and notation are pointed out. It is clarified that the goal is to show 6 divides 7n - 1, not 7n - 6. A correct formulation using the binomial theorem and modular arithmetic is suggested, concluding that 7n - 1 is indeed divisible by 6. The conversation emphasizes the importance of precise notation and understanding modular arithmetic in proofs.
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I am attempting to solve the problem to prove 6 is divisible by 7n-6 Is my logic all correct If it is, i can use it on similar problems!

Proof by induction:
Base case 6 is divisible by 71-1 True! We can continue

Step 2: Assume n=k is true, prove k+1 is true.
Since n=k is true, 6m = 7k-1
therefore, 7k = 6m+1

6 | 7k+1 -1
6 | 7 * 7k -1
6 | 7 * 6m + 1 - 1 (substitute 7k for 6m+1)
6 | 6m * 7
Any multiple of 6 must be divisible by 6.

I feel like I am mistaken somewhere, so feel free to correct me if I'm going on the wrong direction,
Thank You!
 
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TinaSprout said:
I am attempting to solve the problem to prove 6 is divisible by 7n-6 Is my logic all correct If it is, i can use it on similar problems!

Hey Tina! ;)

Your general approach is correct.
It's just that I see a couple of mistakes in there.

For starters, isn't it '6 divides 7n-1' that you want to prove instead of '6 is divisible by 7n-6'?

TinaSprout said:
Proof by induction:
Base case 6 is divisible by 71-1 True! We can continue

Step 2: Assume n=k is true, prove k+1 is true.
Since n=k is true, 6m = 7k-1
therefore, 7k = 6m+1

6 | 7k+1 -1

This is what we want to prove, so we cannot write it as if we already know it's true.
Instead we should write '7k+1 -1' and try to rewrite this expression such that is becomes obvious that it's divisible by 6.

TinaSprout said:
6 | 7 * 7k -1
6 | 7 * 6m + 1 - 1 (substitute 7k for 6m+1)

I'm missing a couple of parentheses here when doing the substitution.
Shouldn't it be 7 * (6m + 1) - 1? (Wondering)

TinaSprout said:
6 | 6m * 7
Any multiple of 6 must be divisible by 6.

I feel like I am mistaken somewhere, so feel free to correct me if I'm going on the wrong direction,
Thank You!

What we should have is:
$$7^{k+1} -1 = 7 \cdot 7^k -1 = 7 \cdot (6m + 1) - 1 = 7\cdot 6m + 6 = 6(7m + 1)$$
Since this is divisible by 6, it completes the induction step.

Therefore 6 divides 7n-1.
 
Another approach would be to use the binomial theorem:

$$7^n-1=(6+1)^n-1=6\sum_{k=1}^{n}\left({n \choose k}6^{k-1}\right)$$
 
MarkFL said:
Another approach would be to use the binomial theorem:

$$7^n-1=(6+1)^n-1=6\sum_{k=1}^{n}\left({n \choose k}6^{k-1}\right)$$

Ah, but we can also calculate modulo 6 (which admittedly relies on the binomial theorem):

$$7^n - 1 \equiv 1^n -1 \equiv 0 \pmod 6$$
 
I like Serena said:
$$7^n - 1 \equiv 1^n -1 \equiv 0 \pmod 6$$

Hi I like Serena. Could you give a detailed explanation?
 
greg1313 said:
Hi I like Serena. Could you give a detailed explanation?

It's part of Modular Arithmetic.

We write $a\equiv b \pmod n$ if there is some integer $k$ such that $a=b+kn$.
That is, $a$ and $b$ are the same except possibly for a multiple of $n$.

This equivalency has a number of properties.
The most notable ones are:

$$a \equiv 0 \pmod n \quad\Leftrightarrow\quad n \mid a \qquad \text{(that is, }n\text{ divides }a\text{)}$$

$$a + b \equiv (a \bmod n) + (b \bmod n) \pmod n$$

$$a \cdot b \equiv (a \bmod n) \cdot (b \bmod n) \pmod n$$

$$a^m \equiv (a \bmod n)^m \pmod n$$

$$a^m \equiv a^{m \bmod \phi(n)} \pmod n\quad$$ if $\gcd(a,n)=1$ and where $\phi(n)$ is Euler's totient function.There are more advanced theorems like the Chinese Remainder Theorem and the Quadratic Residue with the corresponding Legendre symbol $\left(\frac ap\right)$.
But those are quite a bit harder to grasp.
 
Ah, so in general $(x^k-1)\equiv0\pmod{x-1}$.
 
greg1313 said:
Ah, so in general $(x^k-1)\equiv0\pmod{x-1}$.

Yep! (Nod)

(And I guess you've noticed that $\LaTeX$ is chockful with support for modulo operations like \pmod and \bmod. ;))
 
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