How to Convert a Positive Integer in Modular Arithmetic - Cryptography - Lesson 3
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- Опубликовано: 11 окт 2024
- In this video, I explain how to convert a positive integer to a congruent integer within a given modulo.
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After many lectures and having read my text book numerous times, I finally understand this connotation of modulo. Thank you so much
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.
Except that negatives are handled differently and only the one's place is represented by the mod.
No one could be more clear...would be the luckiest to attend ur math classes 🤩🤩🤩🤩😇
finally understood the modular arithmetic concept.thanks
OMG im watching in 2021 and i have to say good job loosening the difficulty of the topic for me i really appreciate
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The universe has led me to this video. Thanks sir
this is a tough topic for me you make it easy thanks a lot will you please explain the pigeon hole principle
these videos are clarifying so much, thank you thank you!
ikr
Perfect explanation
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
wow you explain soo good
thank you,.
Thank you sir.
Thank you so much.
Thanks aton
Thank you, it helped me!
Thank you
thank you thank you thank you
How would I do it with a number too large to calculate? Say 3^3483 or something similar? (loving the videos mate)
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.
Thank You!!
You are a God, sir! Appreciate
THANK YOU
Nice work
Thank U Sir !!
Bravo
Tq
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?
Encryption.
I am shocked that I got that. 😯
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
Can you do this for arbitrary non-integer numbers?
go to 0:35 and tap the 1 key continously and see what it sounds like
What? How is 3/4 zero with remainder 3?
That is why I need to now if someone can help here
@@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
Finally some one asked genius 😂 ❓
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@@rechargeablefan is that Japanese for what? Nani?
Yes
@@rechargeablefan Cool. You taught me something as well! 👍🏼
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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
SirXavior
2 is the R after fing out all thevfull number of times that 3 can fit in 5.... So it works fine :)
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.
+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
3 and 5 have the same remainder when divided by 2, so they're congruent by transitivity
7 mod 4
Good work