Prove that

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

is divisible by 7 for all positive integers \(n\).

Prove that

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

is divisible by 7 for all positive integers \(n\).

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TopNewestWe 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. – Adarsh Kumar · 2 years, 8 months ago

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– Sharky Kesa · 2 years, 8 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.Log in to reply

– Adarsh Kumar · 2 years, 8 months ago

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

– Sharky Kesa · 2 years, 8 months ago

Sharky Kesa is a pseudonym. Sort of like Robert Saunders - Benjamin Franklin.Log in to reply

– Krishna Ar · 2 years, 8 months ago

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

– Aditya Raut · 2 years, 8 months ago

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

@Adarsh Kumar -You a telugu? – Krishna Ar · 2 years, 8 months ago

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

– Adarsh Kumar · 2 years, 8 months ago

no i am not a telugu.I am a northie.Log in to reply

– Krishna Ar · 2 years, 7 months ago

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

– Adarsh Kumar · 2 years, 7 months ago

nopesy i study in a government school.Log in to reply

– Krishna Ar · 2 years, 7 months ago

Seriously???/Then how you so intelligent?Log in to reply

– Adarsh Kumar · 2 years, 7 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.Log in to reply

– Sharky Kesa · 2 years, 8 months ago

Fortunately, I can read that. Do you have a Hindi keyboard?Log in to reply

– Aditya Raut · 2 years, 8 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..Log in to reply

– Sharky Kesa · 2 years, 8 months ago

OK. Did you understand my Hindi I typed up there?Log in to reply

– Aditya Raut · 2 years, 8 months ago

indi ?Log in to reply

– Sharky Kesa · 2 years, 8 months ago

Hindi. Sorry.Log in to reply

@Sharky Kesa in how many things are you so much expertise attainer !!! – Aditya Raut · 2 years, 7 months ago

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

– Sharky Kesa · 2 years, 7 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.Log in to reply

– Aditya Raut · 2 years, 7 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"Log in to reply

– Sharky Kesa · 2 years, 7 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.Log in to reply

– Aditya Raut · 2 years, 7 months ago

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

– Sharky Kesa · 2 years, 7 months ago

sharkesa@gmail.com, what's yours?Log in to reply

– Aditya Raut · 2 years, 7 months ago

I am a volleyballer :P What sport u play ?Log in to reply

– Sharky Kesa · 2 years, 7 months ago

Cricket. :DLog in to reply

– Adarsh Kumar · 2 years, 8 months ago

yeah sorry i posted it in a haste.Log in to reply

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

img

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– Sharky Kesa · 2 years, 8 months ago

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

knowyouknowbut what i wanted toknowis whether youknowwhat the person in imageknows about what youknow,knowit ? – Aditya Raut · 2 years, 8 months agoLog in to reply

– Sharky Kesa · 2 years, 8 months ago

know-ception.Log in to reply

– Daniel Liu · 2 years, 8 months ago

\(50^n+48\times 99^{3n-1}\equiv 1^n+-1\times 1^{3n-1}\equiv 0\pmod{7}\)Log in to reply

– Sharky Kesa · 2 years, 8 months ago

The solution I was looking for.Log in to reply

– Kartik Sharma · 2 years, 7 months ago

I thought the same!!Log in to reply

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, \(50^n+48\times99^{3n-1}\equiv 1^n+(-1)\times1^{3n-1}\equiv 0\pmod {7}\) – Mas Mus · 2 years, 2 months ago

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