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.

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.

@Sharky Kesa
–
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..

@Aditya Raut
–
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.

@Sharky Kesa
–
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"

@Aditya Raut
–
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.

@Krishna Ar
–
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.

$</code> ... <code>$</code>...<code>."> Easy Math Editor

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boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

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## Comments

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

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

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I

knowyouknowbut what i wanted toknowis whether youknowwhat the person in imageknows about what youknow,knowit ?Log in to reply

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

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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.

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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.

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

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

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

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

Oh my goddd !!!!!Log in to reply

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@Adarsh Kumar -You a telugu?

Wow! It did ring a bell but yeah..we Southies aren't much used to it :pLog in to reply

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yeah sorry i posted it in a haste.

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

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