Induction Proof that 7^n - 1 is Divisible by 6

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We see an easy divisibility proof using induction. Mathematic induction is a tremendously useful proof technique and today we use it to prove that 7^n - 1 is divisible by 6. This is an exercise from Introductory Discrete Mathematics by V.K. Balakrishnan.

Thanks to Nasser Alhouti, Robert Rennie, Barbara Sharrock, and Lyndon for their generous support on Patreon!

I hope you find this video helpful, and be sure to ask any questions down in the comments!

+WRATH OF MATH+

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Thank you very much sir.
It is even the exact question that is in my handout my lecturer gave us that we couldn't understand what he solved. 😊 Thanks

CovenantOluwaferanmi

Thank you so much for the well-explained video!

jacksongraham

I mean (a+1)^n-1 = 0 mod(a) proof is trivial using binomial and (a-1)^2n-1= 0 mod(a) if i am not mistaken albo trivial by binomial extension. i found this channel recently and i find it really awesome, greetings from Poland

szymon

Thank you so much sir for the video. My answer came right.

atharvavlogs

Got this one X^(2n -1) + Y^(2n -1) is divisible by (x + y)

revitech

I actually solved it when I tried it myself at 1:00 QuQ
There is hope for me.

jboj

INDIAN BOOK amazing
love you
I am from India

SHASHANKRUSTAGII

You could also use the factorization a^n - b^n = (a - b)(a^(n-1) + ... + b^(n-1)) (using a=7, b=1) or the binomial theorem 7^n = (6 + 1)^n = 6^n + n*6^(n-1) + ... + n*6 + 1. Of course, these could (and perhaps should) be established using induction.
On a separate note, some of us consider 0 to be a natural number. It might depend on whether we're referencing human nature or Mother Nature.😁

tomkerruish

how do you proof it for all integers n?

vindookiediesel

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