How to Convert a Positive Integer in Modular Arithmetic - Cryptography - Lesson 3

[@jonhorn5141]

After many lectures and having read my text book numerous times, I finally understand this connotation of modulo. Thank you so much

[@austingriner8179]

This is practically the same thing as counting in a different base. I.e. mod 2 is binary, mod 10 is decimal, mod 12 is duodecimal, mod 16 is hexadecimal, etc.

[@austingriner8179]

Except that negatives are handled differently and only the one's place is represented by the mod.

[@artellasentertainment9634]

No one could be more clear...would be the luckiest to attend ur math classes 🤩🤩🤩🤩😇

[@thobokgari8062]

finally understood the modular arithmetic concept.thanks

[@elitemaths4994]

East or West, videos with white backgrounds and human handwriting are the best !!!

[@dilnasha.m9807]

faaacts

[@don_1019]

OMG im watching in 2021 and i have to say good job loosening the difficulty of the topic for me i really appreciate

[@joeandcoffee]

The universe has led me to this video. Thanks sir

[@Himynameisoleg_]

these videos are clarifying so much, thank you thank you!

[@aydenmudge279]

ikr

[@C4d009]

Thank you! Grazie! Danka! Arigato! Obrigado! Shookram! Kapun Kap! Merci! Umgoi! Gracias! Tak!

[@binyamin4649]

this is a tough topic for me you make it easy thanks a lot will you please explain the pigeon hole principle

[@oceanlight9329]

Nice vid gj :) Also I’d recommend using a crosshair pointer if possible - I think it looks a bit more professional, as well as making things more clear :p

[@ReniChris07]

Perfect explanation

[@detroitstudios397]

go to 0:35 and tap the 1 key continously and see what it sounds like

[@danielgaray338]

wow you explain soo good

[@gkarthikraja1072]

Thank you, it helped me!

[@bernatbosch]

If several numbers have the same congruencies, once you've got the congruent number and the mod you cannot know what number was at the origin. What's the use of this?

[@TripoliLy]

Encryption.

[@mychannelofawesome]

Can you do this for arbitrary non-integer numbers?

[@athosbrocker9785]

You are a God, sir! Appreciate

[@Coach_MarcyBob]

I am shocked that I got that. 😯

[@haucaidevil2703]

so the number before the (mod n) has less value than n? your other video says 3 = 5 (mod 2) and you explain it in another way

[@funforeveryone13]

How would I do it with a number too large to calculate? Say 3^3483 or something similar? (loving the videos mate)

[@learnmathtutorials]

Great question. Basically, you would want to convert your exponent to base 2 (binary) then perform modular binary exponentiation to get your answer. I will make sure to create a video explaining how to do this and put it into the Cryptography playlist.

[@kevincoyle8088]

Thank you so much.

[@a.sathyajothi2580]

Thank you sir.

[@ThomasArlen-u5l]

Brennan Shore

[@leonardobarrera2816]

What you sell?

[@kaankayis6057]

Thank You!!

[@navanshu]

Good work

[@omkarmarbhal]

Thank U Sir !!

[@geetu291]

Thank you

[@Victual88]

Thanks aton

[@sazsaiglesias]

thank you,.

[@elianpulido3360]

THANK YOU

[@prawnsean]

thank you thank you thank you

[@KingofArmageddon20]

What? How is 3/4 zero with remainder 3?

[@hmd9653]

That is why I need to now if someone can help here

[@riddhimanna8437]

@@hmd9653 3 = 4*0 + 3 (If the dividend is smaller than the divisor then multiply the divisor with 0 or a negative number to get as close to the dividend as possible but lesser than the dividend then you get the smallest positive remainder: some more egs: 9/17 ...9=17*0 + 9, -3/4.... -3 = 4*(-1) + 1

[@uchavrittibhagavatars5736]

Finally some one asked genius 😂 ❓

[@razanrzk]

Tq

[@sanelprtenjaca9776]

Bravo

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[@rechargeablefan]

This dude has not been uploading for 6 years.

[@learnmathtutorials]

Yet he lurks in the comments ... 😏

[@rechargeablefan]

NANI!?!?

[@learnmathtutorials]

@@rechargeablefan is that Japanese for what? Nani?

[@rechargeablefan]

Yes

[@learnmathtutorials]

@@rechargeablefan Cool. You taught me something as well! 👍🏼

[@sirxavior1583]

Hmm.. Doesn't work for the last example in the previous video 3= 5 (mod 2). 3/2 = 1r 5 3 = 2 * 1 + 5 = 2 + 5 3 != 7

[@fredi1356]

SirXavior 2 is the R after fing out all thevfull number of times that 3 can fit in 5.... So it works fine :)

[@artads8745]

i think the reminder is 1. is'nt it? 3/2=1r 1 3 = 2 * 1+1 3=2+1 please correct me if am wrong, thank you.

[@merakiday1628]

+Art Ads You could try: 3=5(mod 2) which can be written 2|3-5=k 3-5=2k 3=2k+5 where k is an interger, and 5 is the remainder

[@sireevictineerivero342]

3 and 5 have the same remainder when divided by 2, so they're congruent by transitivity

[@nabeeliqbal337]

7 mod 4

[@SzTz100]

Nice work