Modulo Challenge

Prove that

50n+48×993n150^n + 48 \times 99^{3n - 1}

is divisible by 7 for all positive integers nn.

Note by Sharky Kesa
5 years ago

No vote yet
1 vote

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

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

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

img img

Aditya Raut - 5 years ago

Log in to reply

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

Sharky Kesa - 5 years ago

Log in to reply

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 ?

Aditya Raut - 5 years ago

Log in to reply

@Aditya Raut know-ception.

Sharky Kesa - 5 years ago

Log in to reply

@Sharky Kesa 50n+48×993n11n+1×13n10(mod7)50^n+48\times 99^{3n-1}\equiv 1^n+-1\times 1^{3n-1}\equiv 0\pmod{7}

Daniel Liu - 5 years ago

Log in to reply

@Daniel Liu The solution I was looking for.

Sharky Kesa - 5 years ago

Log in to reply

@Daniel Liu I thought the same!!

Kartik Sharma - 5 years ago

Log in to reply

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.

Adarsh Kumar - 5 years ago

Log in to reply

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 - 5 years ago

Log in to reply

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

Adarsh Kumar - 5 years ago

Log in to reply

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

Sharky Kesa - 5 years ago

Log in to reply

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

Krishna Ar - 5 years ago

Log in to reply

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

Aditya Raut - 5 years ago

Log in to reply

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

Sharky Kesa - 5 years ago

Log in to reply

@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 - 5 years ago

Log in to reply

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

Sharky Kesa - 5 years ago

Log in to reply

@Sharky Kesa indi ?

Aditya Raut - 5 years ago

Log in to reply

@Aditya Raut Hindi. Sorry.

Sharky Kesa - 5 years ago

Log in to reply

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

Aditya Raut - 5 years ago

Log in to reply

@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 - 5 years ago

Log in to reply

@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 - 5 years ago

Log in to reply

@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.

Sharky Kesa - 5 years ago

Log in to reply

@Sharky Kesa I am a volleyballer :P What sport u play ?

Aditya Raut - 5 years ago

Log in to reply

@Aditya Raut Cricket. :D

Sharky Kesa - 5 years ago

Log in to reply

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

Aditya Raut - 5 years ago

Log in to reply

@Aditya Raut sharkesa@gmail.com, what's yours?

Sharky Kesa - 5 years ago

Log in to reply

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

Krishna Ar - 5 years ago

Log in to reply

@Krishna Ar no i am not a telugu.I am a northie.

Adarsh Kumar - 5 years ago

Log in to reply

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

Krishna Ar - 5 years ago

Log in to reply

@Krishna Ar nopesy i study in a government school.

Adarsh Kumar - 5 years ago

Log in to reply

@Adarsh Kumar Seriously???/Then how you so intelligent?

Krishna Ar - 5 years ago

Log in to reply

@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.

Adarsh Kumar - 5 years ago

Log in to reply

yeah sorry i posted it in a haste.

Adarsh Kumar - 5 years ago

Log in to reply

We will show that 50n+48×993n10(mod7)50^n+48\times99^{3n-1}\equiv 0\pmod {7}

We have 501(mod7)50\equiv 1\pmod {7}, 481(mod7)48\equiv -1\pmod {7}, and 991(mod7)99\equiv 1\pmod {7}

Then, 50n+48×993n11n+(1)×13n10(mod7)50^n+48\times99^{3n-1}\equiv 1^n+(-1)\times1^{3n-1}\equiv 0\pmod {7}

Mas Mus - 4 years, 7 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...