Thank you so much for your explanation...really means a lot.... because of this vedio I got 14 marks in my semester exam....thank you so much.... worth watching it👍
i m 15 and i am super crazy about coding...ethical hacking etc....but other videos on youtube are super complicated for me to understand...but thank you so much brother for making videos too easy to understand❤🤙😃
'e' ki value ke liye zaruri hai.. ki highest cofactor lena hai?? ya fir koi bhi co factor le saktehai??.... like 4:59 yaha pe 7 ki jagah 9 le sakte hai kya??
Thank U so much bhaiya. Ek question hai encryptex text ka koi constraint hai kya... Jaise encryptex text p and Q dono se chota hona chahiye ya waisa kuch?
brother 4th step me e ki value 7 kesy le hai ? koi aur value le ksty hain agr me 3 ln tw 3 ( 20 se choti hai aur 1 se bare ) aur gcd (3,20)=1 bhi hai tw 7 kesy le hai?
Doubt :- encryption M(e) mod n and public key is (e,n) - (3,33 ) then how you take value of (e) 31 becoz ydi mujhe kisi ki public key se encryption krna ho to mujhe to sirf uski public me jo diya he usi se krna pdegi ...usne public key kese banayi kesi ko thodi pata rhegi
Jo e ki value aapne assume ki hai wha pr 3,7,9,11,13,17 inme se kuch v aa skta hai agr hum kuch aur lete hai to ans change nhi hoga aur kya wo tb v shi mana jayega??
sir jaisa ka ap na bataya ka e d mod Φ(n)=1 ma sa e ka aisa multiple banayain ka mod uska 1 remainder da, but sir what will answer if e is "35" and Φ(n) is 192 then sir its not possible because multiple of 5 is 35 * 5 175 and 35 * 6 is 210 so in this case its not possible...and my main question is this, should we implement in all scenarios?
In RSA if the given plaintext is "hi" then what should we use while doing encryption in the place of M (plaintext). Please say me the process as soon as possible
Sir could you please explain how you got 31^7mod(33)....i tried using remainder theorem which was available online when i applied it i am not arriving to the answer 4 could you please explain regarding this
i am also not getting the answer it should be 15 31^7 mod 33 31^2 mod 33 it will be 6 so we will go for 31^4 it will be square of 6 then mod 33 so it will come 36 mod 33 --> 3 then again 31^2 ---> 6 and 31^1 -->31 so the answer we got should be 31^4 *31^2 *31^1 mod 33 answer need to 15
31 mod 33 =-2 reminder, (-2)^7 mod 33 = (-2) ^5 * (-2)^2 mod 33 (-2) ^5 mod 33 * (-2)^2 mod 33 1 * 4 =4 note : 31 mod 33 =31 but can be written as 31-33=-2 reminder , the reminder property
@Chirag Bhalodia but we saw that in rsa algorithm genrate to type key one is public and second is pravite and in this algorithm public key does exchange it's means in this algorithm we use also deffi hell mam algorithm please make video on it in which explain this thing when we use rsa that time how will exchange public key
Thank you so much for your explanation...really means a lot.... because of this vedio I got 14 marks in my semester exam....thank you so much.... worth watching it👍
14/100 🐈
@@billionairediarie 1 unit 14 ka aata hai bhai
This helps me during my exam.Thank you so much❤️
i m 15 and i am super crazy about coding...ethical hacking etc....but other videos on youtube are super complicated for me to understand...but thank you so much brother for making videos too easy to understand❤🤙😃
Welcome 😊✌️
Bhai coding mat le, bohot tough hai
me also bro
I m just 10 year old n m loving coding
I Am just born now and love coding
Hi bro thanks a lot because of ur video only i could understand rsa algorithm properly and i completed my diploma thanks a lot for ur help 🎉🥳😊👏
Can u help me with one question?
1. What mathematics courses or topics do I need to learn to understand the alogrithm
Bahut bahut dhanyawad guruji🙏❤️😇 really appreciating job done by u😇
excellent explaination each and every point
thanks sir
sir apki wja se mjy netwrk security subj aya hai thnku so much sir ...love from pakistan 💜
best playlist of cryptography 🙏 , thank you soo much sir
Thank you, watching this a night before exam and its very helpful 🔥
Thanks
no other channel could explain better than you 🙏
Thank you so much @Abhishek Sharma
Anna tumba dhanyawadagalu ...thanks a lot💯🔥
Thank you so much sir for this
Aap bahut acha pdate ho...
Sb smz mai aajata h 🥰🥰🥰🥰
In a public-key system using RSA, you intercept the ciphertext C = 20 sent to a user, whose public key is e=13 n = 77. Calculate the plaintext M.
for finding d=1+k(phi)/e where k=0,1,2,... and d is not in decimal point.
Thankyou so much this is very useful for me today is my exam and it is easy to understand 👌👌
bestt video sir thanku i m a bca student
My Sir watches your videos. Big fan
'e' ki value ke liye zaruri hai.. ki highest cofactor lena hai?? ya fir koi bhi co factor le saktehai??.... like 4:59 yaha pe 7 ki jagah 9 le sakte hai kya??
Thank U so much bhaiya.
Ek question hai encryptex text ka koi constraint hai kya... Jaise encryptex text p and Q dono se chota hona chahiye ya waisa kuch?
Brooo...😉😉💕💕,Thank u so much u helped me in Exam
brother 4th step me e ki value 7 kesy le hai ? koi aur value le ksty hain agr me 3 ln tw 3 ( 20 se choti hai aur 1 se bare ) aur gcd (3,20)=1 bhi hai tw 7 kesy le hai?
Doubt :- encryption M(e) mod n and public key is (e,n) - (3,33 ) then how you take value of (e) 31 becoz ydi mujhe kisi ki public key se encryption krna ho to mujhe to sirf uski public me jo diya he usi se krna pdegi ...usne public key kese banayi kesi ko thodi pata rhegi
Sir you are amazing I watch last Saturday your. Lecture about Des it's very helpful for us Amazing ❤
From where you take 33 in public and private key?
Your video's are so much helpful .. great job
This content is on par with one of the best videos available on this topic nice work
Thanos chup baithh jyada mat bol 😡
Engg lelo beta apne ap sab gayab hogaa so called interest😂😂😂😂
I also Love to make things simple ❤💯
Very good channel for better understanding
very great and clear explanation thank you bhai
Thank you so much
Very good yar 👌🏻
@3:30, please explain it in detail--> ed MOD phi(n) = 1
Very good explanation sir thank you so much it's very helpful for me sir thanks again 🙏
best explanation with super flow...thanks alot
Jo e ki value aapne assume ki hai wha pr 3,7,9,11,13,17 inme se kuch v aa skta hai agr hum kuch aur lete hai to ans change nhi hoga aur kya wo tb v shi mana jayega??
Thanks sir ❤
Thanks a lot sir this video really helpful to understand the concept
Thank you so much❤😊👍
This is best video for rsa.
good explanation
Mauj kar di bhai...
Awesome explanation 🔥🔥🔥
Let a, b, and m be integers. If m ≥ 2 and a ≡ b (mod m), then gcd(a, m) = gcd(b, m). Can u solve this by RSA
yes
Thank you sir.. Your videos are very helpful for me..Keep it up🙏🙏
Excellent explanation thanks sir...
nice explanation of RSA algorithm bro
very nice explanation bro.
Bhai thxxx , bhot acha explain kiya ❤❤❤
superb video
Sir Elgamal algorithm k lecture bh upload kr den as soon as possible
i dnt understood single word of hindi but I understood your concept perfectly.. thank you
Very Informative class. Very easy to understand
sir jaisa ka ap na bataya ka e d mod Φ(n)=1 ma sa e ka aisa multiple banayain ka mod uska 1 remainder da, but sir what will answer if e is "35" and Φ(n) is 192 then sir its not possible because multiple of 5 is 35 * 5 175 and 35 * 6 is 210 so in this case its not possible...and my main question is this, should we implement in all scenarios?
thank you sir its very helpful 🙏
Kaphi Acha explain Kar raha hai bro, keep going
Amazing explain appreciated 😊
sir, in your example, while choosing 'e' can it be 3 also, when we will find gcd(20,3) it will give output as 1.
or else we can take any value?
There will always be 2 values you can choose any value because d is the multiplicative inverse of e and both values will be chosen.
@@sameerkarur6624 ty 😀
@@sameerkarur6624 but when I tried with e=3 we will get the wrong answer
So please can u help me out to figure out how to select e
You explained it in very present and easy manner I like it.
u r way better than my teacher
In RSA if the given plaintext is "hi" then what should we use while doing encryption in the place of M (plaintext). Please say me the process as soon as possible
We can only take integers as the input.
Write H=8 and I= 9 and in place of M place 89
Bahut badhiya explanation.
Bhai you are a legend❤❤❤❤❤
Thanks a lot bhiya 🙏🙏🙏
Very nice explained
Thankuu so much sir, for explaining this topic in a simple manner.🙏🙏
I'm studying MCA,
I have paper of information and network security on tomorrow
So can you help me during exam
we can also use e=3 as hcf of (20,3) is also 1?
yes
Very nice and clear explanation 👍
Thank you so much sir
Thank u so much sir ... Your video is Very helpful for me...
thank you friend
Achcha Samjhaya Bhai.
Thanks man
Thank you sir
Amazing explanation .... ❤❤❤
when abishek is here why fear thanks abhishek i am doing this before the day of my
exam
Bhai tumara help kabhi nahi bhisareyga today my exam wrote well it is u r pleasure bhai ❤️❤️
Ye to understand h, ye to too good h 😃
Great presentation of Concepts.
Thank You sir!!
Perfect explanation. Thank u sir
can we take M=100 for the same example... ?
e nikalte waqt p and q mein se koi ek value le sakte hai dono hi 1 se bade aur 20 se chote hai?
Great sir👍
Very helpful brother 👌👏👏👏❤
Good explanation...
Bhai e p and q se different and 1 and fi(n) ke beech me hona chahiye?
Thanks alot man. You are a life saver!
aag laga di sir aag
Good explanation and also given pre requisite in description 💯
Sir could you please explain how you got 31^7mod(33)....i tried using remainder theorem which was available online when i applied it i am not arriving to the answer 4 could you please explain regarding this
Same... Did u got ur answer?
i am also not getting the answer it should be 15
31^7 mod 33
31^2 mod 33 it will be 6 so we will go for 31^4 it will be square of 6 then mod 33 so it will
come 36 mod 33 --> 3
then again 31^2 ---> 6 and 31^1 -->31
so the answer we got should be 31^4 *31^2 *31^1 mod 33
answer need to 15
31 mod 33 =-2 reminder,
(-2)^7 mod 33 = (-2) ^5 * (-2)^2 mod 33
(-2) ^5 mod 33 * (-2)^2 mod 33
1 * 4 =4
note : 31 mod 33 =31 but can be written as 31-33=-2 reminder , the reminder property
If A is sender of messenger then B should be receiver and how will B get private key of A to decrypt?
Sir plz solve example on it we have to prepare for exam.
Really appreciable work done by you🙌👏
AKTU student👩🎓
Thsnkuu so much 💞💞💞sir Auesome lecture
Amazingly explained THanks a lot !!!
@Chirag Bhalodia but we saw that in rsa algorithm genrate to type key one is public and second is pravite and in this algorithm public key does exchange it's means in this algorithm we use also deffi hell mam algorithm please make video on it in which explain this thing when we use rsa that time how will exchange public key
excellent. make more video like that.