Prove that (3n+1)7^n - 1 is divisible by 9 (Mathematical Induction)

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proof of sum-of-digits = multiple of nine given base is a multiple of 9:
each digit M represents some M*10^k, but {10 -1 is a multiple of 9 and 10^(k+1)-1=9*(10^k)+10^k-1} [inductive proof within a non-inductive proof], therefore (M*10^k)%9=M for all integer M where 0<M<9, and =0 if M=9, but an improper definition of mudulo with the property 9%9=9 will be used, which does not disturb the divisibility.
summing all digit values is equivalent to taking the sum of the modulos of the digits, which by the properties of modulo preserves the remainder and thereby the divisibility or lack thereof of any number. thus if the sum of the modulos is a multiple of 9, the original was also. if the sum of the modulos is a multi-digit value, repeat the process until 9 or 0 is found.

MrRyanroberson

great video honestly learned more from this than 3 weeks of my lecture combined

jackzhang

ok, now i saved one right for my self on the exam!!!!

bethelhemsolomon

i was stuck on his question for a good 20 minutes and my book looks like a crime scene and you tell me to sub in 1 ;_;

jingjiezheng

wow wow well explained you nailed it man

philasandegoodwill

Why is it +9m in the proof? While in assumption is 1= (3k+1)7'k -9m. Im just little bit confused.

joyleemorrondoz

this is way easier to understand than the hint that i got from my teach -.-

happylunch

Here U want a lil bag I’m about to go down! Wrong you

sunrain

Prove that (3n+1)7^n - 1 is divisible by 9 (Mathematical Induction)

Prove that (3n+1)7^n - 1 is divisible by 9 (Mathematical Induction)

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