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# Modulo Challenge

Prove that

$50^n + 48 \times 99^{3n - 1}$

is divisible by 7 for all positive integers $$n$$.

Note by Sharky Kesa
3 years, 2 months ago

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We know that (a+b)(mod m)=a(mod m)+b(mod m) and (ab)(mod m)=a(mod m) x b(mod m) after knowing this it is a child's play. 50=1(mod 7) that implies 50^n=1(mod 7) now keep that aside 48=-1(mod 7) and 99=1(mod 7) that implies 99^(3n-1)=1(mod 7) that implies [99^(3n-1) x 48]=-1(mod 7) Adding both these equations we get 50^n +[48 x 99^(3n -1)]=0(mod 7) done done done dunna done. P.S:THE LAST REMARK WILL ONLY BE UNDERSTOOD BY INDIANS.

- 3 years, 2 months ago

I can understand it , but ten again, I am Indian. Me re ko patha ha hindi likin me re pas nahi ha hindi keyboard. That's me writing hindi, hope you can understand.

- 3 years, 2 months ago

wow in never knew that u were INDIAN .PRETTY SURPRISING!!

- 3 years, 2 months ago

Sharky Kesa is a pseudonym. Sort of like Robert Saunders - Benjamin Franklin.

- 3 years, 2 months ago

Very good solution..I too had the same one..."The last comment was not really guess-able"...ELucidate?

- 3 years, 2 months ago

Done dunna done.... is actually " डन डना डन " which is synonym for a ecstatic way of saying "it's done"

- 3 years, 2 months ago

Wow! It did ring a bell but yeah..we Southies aren't much used to it :p @Adarsh Kumar -You a telugu?

- 3 years, 2 months ago

no i am not a telugu.I am a northie.

- 3 years, 2 months ago

OH..Well :)..You study in one of the Narayana /Chaiatanya schools?

- 3 years, 2 months ago

nopesy i study in a government school.

- 3 years, 2 months ago

Seriously???/Then how you so intelligent?

- 3 years, 2 months ago

in maths particularly number theory i learned it all by myself mainly from brilliant.but other math topics were taught to me by my father.in science i am 0.

- 3 years, 2 months ago

Fortunately, I can read that. Do you have a Hindi keyboard?

- 3 years, 2 months ago

I used the google translate, that is to input hindi there we have handwritten or even phonetic keyboard. btw i have a software too, it's called "BarahaPad" ,almost all indian languages can be typed by it..

- 3 years, 2 months ago

OK. Did you understand my Hindi I typed up there?

- 3 years, 2 months ago

indi ?

- 3 years, 2 months ago

Hindi. Sorry.

- 3 years, 2 months ago

Oh my goddd !!!!! @Sharky Kesa in how many things are you so much expertise attainer !!!

- 3 years, 2 months ago

In how many things are you so much what? I don't get what you mean by expertise attainer. I live in a very English background. If some sentence doesn't make sense to me, there's probably something wrong. BTW, attainer isn't a word.

- 3 years, 2 months ago

Haha we had that named section in our PACE booklets, it's the exercise which is the toughest in the whole book :P By saying to you i meant simply "expert"

- 3 years, 2 months ago

I am an expert mainly in math and science (though not really). Calvin Lin is an expert at math. I am nowhere near him. I am alright in Hindi, but exceptional in English. I am above average in sport. Above national average in endurance. But yeah, I am a all-rounder you could say.

- 3 years, 2 months ago

I think these talk can better be done on google or email..... may i have your email ?

- 3 years, 2 months ago

sharkesa@gmail.com, what's yours?

- 3 years, 2 months ago

I am a volleyballer :P What sport u play ?

- 3 years, 2 months ago

Cricket. :D

- 3 years, 2 months ago

yeah sorry i posted it in a haste.

- 3 years, 2 months ago

Modulo challenge is child's play by Induction....

img

- 3 years, 2 months ago

Yes, I have it relatively low levelled. I want people to use modulo. I already know there is a solution using induction.

- 3 years, 2 months ago

I know you know but what i wanted to know is whether you know what the person in image knows about what you know ,know it ?

- 3 years, 2 months ago

know-ception.

- 3 years, 2 months ago

$$50^n+48\times 99^{3n-1}\equiv 1^n+-1\times 1^{3n-1}\equiv 0\pmod{7}$$

- 3 years, 2 months ago

The solution I was looking for.

- 3 years, 2 months ago

I thought the same!!

- 3 years, 2 months ago

We will show that $$50^n+48\times99^{3n-1}\equiv 0\pmod {7}$$

We have $$50\equiv 1\pmod {7}$$, $$48\equiv -1\pmod {7}$$, and $$99\equiv 1\pmod {7}$$

Then, $$50^n+48\times99^{3n-1}\equiv 1^n+(-1)\times1^{3n-1}\equiv 0\pmod {7}$$

- 2 years, 9 months ago

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