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Thanks for ur explanation Sir.Please evaluate my answers once. 1.private key (77,119) , cipher 41 2.private key (11,24) , cipher 18 3.private key (23,187), cipher 146 4,as plain test which is in binary is greater than n value This can't be done. 5.private key (91,221) cipher 181
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How can I thanj you? I study the course, same textbook. You save my time and effort + better understanding. God bless you. Your effort is so appreciated. I am reading all your blog and watching your TH-cam videos.
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exercise 2 my public key is (11,35) and my private key is (11,35). C=18 which is 2^11mod 35(where 35 is gotten from 7*5) and D=2 (which is gotten from 18^11mod35)
No its wrong. Public key is 11 given in question and private key d = 11. Your cipher text is true. C = 18. This video share with others and subscribe my channel...Thank you in advanced...
@@ChiragBhalodia Sir, thank you for your explanation. In this case, it seems the encryption can be reversed with the same key. 2^11 ≡ 18 (mod 35), 18^11 ≡ 2 (mod 35). In fact choosing others (like 17) also have same effect, 2^17 ≡ 32 (mod 35), 32^17 ≡ 2 (mod 35). I'm bit confused with this.
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exercise 1 my public key is (5,199) and my private key is (77,199). i went into calculating my cipher text with the plain text which was given as 6 so it gave me 15 as my cipher text(C=15). and lastly my D=37
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When finding the value of e wch is between 1 and phi(n) and gcd(e,phi(n))=1, are you not suppose to select the lowest or the smallest prime number wch satisfy the condition ...... why you choose 13 if the gdc(7,120)=1 coz 7 is prime and is the lowest prime number we can use as the encryption key
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Sorry for late reply....!!! To find the private key d, the equation is modified you can check in step no 5 (in explanation).... Please share this video with others..... Please subscribe my channel.... Please follow my blog: www.chiragbhalodia.com/2021/09/rsa-algorithm-with-example.html
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Thanks Sir 😊 My concept are clear like crystal. Sir I want to download these slides. If you feel comfortable please give link in the description for downloading.
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Advantage os RSA are: It is very easy to implement RSA algorithm. RSA algorithm is safe and secure for transmitting confidential data. Cracking RSA algorithm is very difficult as it involves complex mathematics. Sharing public key to users is easy. Disadvntage of RSA are: It may fail sometimes because for complete encryption both symmetric and asymmetric encryption is required and RSA uses symmetric encryption only. It has slow data transfer rate due to large numbers involved. It requires third party to verify the reliability of public keys sometimes. High processing is required at receiver’s end for decryption. Please share with others..... keep supporting..... Please subscribe my channel.... Please follow my blog: www.chiragbhalodia.com/2021/09/rsa-algorithm-with-example.html
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If the sender uses RSS algorithm. So he gets two keys Public and Private. If I want to send someone to public key . By What Method . How will I reach that public key.
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@@ChiragBhalodia 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
Thanks sir, I really understand you lecture, my question is this, can we choose any number for public key if the number is a prime number that fall between the range of p and q
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Thank you so much sharing this video and also thanx to sharing answers of exercises.... I have try myself to find private key "d". But in small numbers I have get the key, but in large number I have confused and not getting key.... Please can you share how to find "d" for large numbers.... Thank you in advanced...
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I don't know if you will see my comment but i hope so. How did you calculate the mod.? I mean how do we calculate P^e mod n ? I don't know how to do it
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That was my typing mistake... It was actually (11 - 1) instead of (12-1). Because equation is like Ф(n) = (p-1) * (q-1). Follow my blog : edu-resources1.blogspot.com/
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yes, whenever you try to find d (private key), always starting value of i=1... Please share this videos with others and *subscribe my channel* ... *Follow my blog:* www.chiragbhalodia.com/2021/09/rsa-algorithm-with-example.html
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For RSA algorithm, your plain text must be in numerical values only. For example, The plain text message (BHAI) encrypted with (RSA) algorithm, the characters of the message are encoded using the values 00 to 25 for letters A to Z. .. Here BHAI = 33. Follow my blog for more content : edu-resources1.blogspot.com/
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I think exercise 2 is wrong. You said that for finding d, you need to do ϕ(n)+1/ e until you get two integer numbers. So how d is 11? Because the ϕ(n) is 24. so if we get like with decimal, we times ϕ(n) which is 24 by 2, until the answer is integer. So i am really confused on how you find d, please help me.
There are 3 method of finding "d", one method is explain here.... remaining 2 methods video will be posted soon... mostly in next week... if you want to check the answer of exercise, blog link is given in description
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Hello Jonathan, 1. e is public key. (We known that as per public key cryptography, public key shared with each other in network by all users). You can encrypt with public key of any user A, then only user A is decrypt the message. No one else in network. Because private key will be calculated using user A's Public key.. 2. If you are think about examination purpose, which is already given in question. If you have any doubts you can ask me.. Share with others and subscribe my channel.... Follow my blog : edu-resources1.blogspot.com/2021/07/rsa-algorithm-rsa-algorithm-explain.html
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You can take any number...This is just example.... Subscribe my channel.... Follow my blog: www.chiragbhalodia.com/2021/09/rsa-algorithm-with-example.html
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Hello Anand, I have solved using modular arithmetic method.... Please share this video with others.... Subscribe my channel.... Follow my blog: www.chiragbhalodia.com/2021/09/rsa-algorithm-with-example.html
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I have find this mod using modular arithmetic theorem... Please subscribe my channel... Keep supporting... Follow my blog: www.chiragbhalodia.com/2021/09/rsa-algorithm-with-example.html
Most IMP for tomorrow examination : All cipher techniques,active and passive attack, des, aes, diffie Hellman, rsa, application of digital signature, kerberos, DSA, public key distribution, hmac, all block cipher modes, ssh, SSL
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In examination Plain text will be given in question....refer questions given in videos......here I am explaining example...So, I have taken plain text = 13.
Most IMP for tomorrow examination : All cipher techniques,active and passive attack, des, aes, diffie Hellman, rsa, application of digital signature, kerberos, DSA, public key distribution, hmac, all block cipher modes, ssh, SSL
Modulus is find using arithmetic modular method...... Please share this video with your friends... Follow my blog: www.chiragbhalodia.com/2021/09/rsa-algorithm-with-example.html
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Contents available on my blog... Thanks for watching..... Please share with others.... Please subscribe my channel... Please follow my blog: chiragbhalodia.com
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Solution of all exercise is posted in blog.
Thank you so much sharing this video and also thanx to sharing answers of exercises....
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Thank you so much sir
Sir , This is the best explanation of RSA amongst all youtube videos
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Respect for you, sir. The way you make complex things so simple to understand is just awesome. Thank you sir for such great contents.
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Great explanation now it's very simple kudos to your great work 🔥🔥🔥🔥🔥
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Such a complex topic explained very well..... Finding "d" is not an easiest task.... Thank u for sharing this kind of video...
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THE BEST explanation on youtube
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Okay now I perfectly understand this algorithm 👍👍👍
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I think it's the best video explaining RSA
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Thanks for ur explanation Sir.Please evaluate my answers once.
1.private key (77,119) , cipher 41
2.private key (11,24) , cipher 18
3.private key (23,187), cipher 146
4,as plain test which is in binary is greater than n value This can't be done.
5.private key (91,221) cipher 181
Exercise one true.
Exercise two false
The true is that private key ={ 11,35 },cipher text =18
Exercise three is true.
Exercise four is true too.
Exercise five also true.
Thank you so much Professor. You are always the greatest in TH-cam.
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RSA is complex topic but very well explained in the this video
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super sir , our teacher share your video for RSA and AES . we really grateful to you.
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You have always been the best teacher to understand the topics very easily 🥰 Thank you sir for being always the best.
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Perfect explanation, Genius
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Perfect explanation with example 👍👌
Thank u akash
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Thank you sooooooo much for your clear explanation. Thank you sir
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Thank you sir u saved me🤩✌️
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Apart from my view below
Great explanation Sir
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Sir, Perfect explanation for RSA algorithm...!
Keep watching
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sir please can you send full setps for exercise 1 upto 5
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Perfect Explanation 👌🏻
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the best explanation
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really i enjoyed this session a lot 💚
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Great. Thank so much Sir.
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Really helpful!!THANK YOU SIR!!
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How can I thanj you? I study the course, same textbook. You save my time and effort + better understanding. God bless you. Your effort is so appreciated. I am reading all your blog and watching your TH-cam videos.
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Awesome blog and awesome video
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Very much Help full... thanks
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👏👏👏well explained
Thanx
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Thank you sir 🙏🤗
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exercise 2 my public key is (11,35) and my private key is (11,35).
C=18 which is 2^11mod 35(where 35 is gotten from 7*5) and D=2 (which is gotten from 18^11mod35)
No its wrong. Public key is 11 given in question and private key d = 11.
Your cipher text is true. C = 18.
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@@ChiragBhalodia Sir, thank you for your explanation. In this case, it seems the encryption can be reversed with the same key. 2^11 ≡ 18 (mod 35), 18^11 ≡ 2 (mod 35). In fact choosing others (like 17) also have same effect, 2^17 ≡ 32 (mod 35), 32^17 ≡ 2 (mod 35). I'm bit confused with this.
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Step:- 3 is a wrong (13-1 )*(11-1)= 12*10= 120 is a right
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Thanks very good😊🎉
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@@ChiragBhalodia
Pls
How calac
Encryption message by calculator
And
Decryption
You can use Laptop Calcualtor or use Exponential theorem.
@@ChiragBhalodia
It is prohibited to use it during the exam
nice explanations with example...👌👍
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Very well explained such a complex topic 👌👌👍
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Keep the good work on...
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Hello Sir, Please explain the steps involved in step 7 to get C = 52. How to perform mod?
Calculation using exponential and modular arithmetic..... Modular is preform using modular arithmetic technique
RESPECT 🙏❤️
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Very useful topic sir.
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Thank you 🙏🙏
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amazing video, thanks for sharing!
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exercise 1 my public key is (5,199) and my private key is (77,199).
i went into calculating my cipher text with the plain text which was given as 6 so it gave me 15 as my cipher text(C=15).
and lastly my D=37
Your private key d = 77 Is true.
Your cipher text is wrong. Cipher text C = 41.
@@ChiragBhalodia c 31 sir
Plain text is 5 and cipher text is 31. N value is 119 .
@@sriganeshbalasubramanian1506 C = 41
@@sriganeshbalasubramanian1506 Plain text = 6 and Cipher text = 41. N = 119
Excellent 👌👌
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Very nicely explained the deeper aspects of RSA algorithm. Good going👍
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When finding the value of e wch is between 1 and phi(n) and gcd(e,phi(n))=1, are you not suppose to select the lowest or the smallest prime number wch satisfy the condition ...... why you choose 13 if the gdc(7,120)=1 coz 7 is prime and is the lowest prime number we can use as the encryption key
You also can use 7...
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@@ChiragBhalodia okey thanks bro
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Well explained👍👌
Thank u
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2:05 , why did you substitute q= with 12 instead of 11 ?
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and also for finding the plaintext, for example if the public key is [18,143], so the plaintext will be 18?
no way... it is wrong...
could you please share your whatsapp number, i could message you there instantly? because i have my final exam tomorrow pls
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Very nice explanation
Thank u kishan
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In exercise 4 -in decryption step you got plain text 26 instead of 59
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Step 5 is kind of unintuitive to me. Does e^-1 = 1/13 or does the notation mean something else in this context? Great video!
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To find the private key d, the equation is modified you can check in step no 5 (in explanation)....
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Nice explanation
Thank u dear
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Thanks Sir 😊 My concept are clear like crystal. Sir I want to download these slides. If you feel comfortable please give link in the description for downloading.
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Excellent 👌
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Sir 52^37 mod 143 value is 141
No its 13.
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Hello sir I would like to ask if what are the advantages and disadvantage of RSA ALGORITHM. Thanks for your reply
Advantage os RSA are:
It is very easy to implement RSA algorithm.
RSA algorithm is safe and secure for transmitting confidential data.
Cracking RSA algorithm is very difficult as it involves complex mathematics.
Sharing public key to users is easy.
Disadvntage of RSA are:
It may fail sometimes because for complete encryption both symmetric and asymmetric encryption is required and RSA uses symmetric encryption only.
It has slow data transfer rate due to large numbers involved.
It requires third party to verify the reliability of public keys sometimes.
High processing is required at receiver’s end for decryption.
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Damn!!!
When I heard your voice
I remembered Kamal hasan 😁(In Dashavataram)
No offence just sharing my view
ohhh..... Thanx a lot dear... Kamal sir is legend... I am not....😀
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Very nice video
Thank u
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Nicely explained
Thank u
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In exercise Q1 when i am computing e inverse mod phi of n with extended Euclidean Theorem i am getting d = 24 rather than 77 why is that ?
If the sender uses RSS algorithm. So he gets two keys Public and Private. If I want to send someone to public key . By What Method . How will I reach that public key.
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@@ChiragBhalodia 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
Sure... I will make it.....
Thanks sir, I really understand you lecture, my question is this, can we choose any number for public key if the number is a prime number that fall between the range of p and q
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A1 sirji
Thank you
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Thank you so much sharing this video and also thanx to sharing answers of exercises.... I have try myself to find private key "d". But in small numbers I have get the key, but in large number I have confused and not getting key.... Please can you share how to find "d" for large numbers.... Thank you in advanced...
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2:06 (12 - 1) != 10
Yes dear, It was my typing mistake....
I have one doubt Why did you choose e=13 not e=7 because 7 is also not divisible by 120
Yu can use e=7.
I don't know if you will see my comment but i hope so. How did you calculate the mod.? I mean how do we calculate P^e mod n ? I don't know how to do it
Hello, Abou Nasma..... I am reading all the comments and also giving the answer of all the coments with in 24 hours.
I will post this method on my blog on tomorrow and share link with you......How it actually works.....By the way.. Here I am writing answer directly because this is the methamatical process... I can't explain that process easily in ppts.... (Follow my blog)
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Is there any video where the solved solutions are posted?
No dear, there is no solution. But you can try your own way to solve. Post answer here. If there is wrong then I will help you to solve out....
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Are you solved this practive example?
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(n) = (13-1 ) * (12-1) = 120? isnt it 12*11 = 132 ?
Thanks for rest of algorithmic explanation
That was my typing mistake... It was actually (11 - 1) instead of (12-1).
Because equation is like Ф(n) = (p-1) * (q-1).
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@@ChiragBhalodia thank you sir for clarification... .. Really appreciate your reply
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I want to know how you Find the Cipher Text, once all the rest is found. i dont understand
share your email id, I will share solution with you tomorrow... how to find cipher text??
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CAN WE TAKE THE VALUE OF E AS 7?
Yes you can... But you ask in terms of examination... Mostly e (means public key) will be provided...
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so the value of i is always 1?
yes, whenever you try to find d (private key), always starting value of i=1...
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thanks ....how can i encrypt and decrypt ----->palintext="simple"
For RSA algorithm, your plain text must be in numerical values only.
For example,
The plain text message (BHAI) encrypted with (RSA) algorithm, the characters of the message are encoded using the values 00 to 25 for letters A to Z. .. Here BHAI = 33.
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thanks for response
b=2
h=8
a=1
i=9
how we will get BHAI=33
yes... thats right...
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Ans= Exercise 1 pub key=(5,119) ..C= 41 ....but hard to get d=? priv key=(d,119)
d=77, C=41 (it is true as per your comment)...
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I think exercise 2 is wrong. You said that for finding d, you need to do ϕ(n)+1/ e until you get two integer numbers. So how d is 11? Because the ϕ(n) is 24. so if we get like with decimal, we times ϕ(n) which is 24 by 2, until the answer is integer. So i am really confused on how you find d, please help me.
There are 3 method of finding "d", one method is explain here.... remaining 2 methods video will be posted soon... mostly in next week...
if you want to check the answer of exercise, blog link is given in description
but i did ϕ(n)+1/ e and the answer of d i get is wrong. if ϕ(n)=24 then i need to reach a point where it will be integer. then how is d 11?
the equation is: ((phi(n) * i) +1) /e
13 was there already what is the need of finding plain text then
Yeah because in some university exam, they ask to again find cipher text.... Because there is one condition for plain text, it must be fulfilled otherwise you will not get correct plain text from cipher text.
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how to find the private key if public key of sender or receiver is know?
It is already explain in video. Please watch again.
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bro in first problem that is 11-1 not 12-1
Yes you are right...It was my typing mistake....
How to find the mod in scientific calculator???
Dear it is mathemtical process.... nit easily find in scientific calculator
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i need solutions of these exercises
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@@ChiragBhalodia thanku
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Why did you select 13 as your e please?
Hello Jonathan,
1. e is public key. (We known that as per public key cryptography, public key shared with each other in network by all users). You can encrypt with public key of any user A, then only user A is decrypt the message. No one else in network. Because private key will be calculated using user A's Public key..
2. If you are think about examination purpose, which is already given in question.
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Exercise 3:
encryption C= 5^7 mod 187 = 146
decryption P= 146^23 mod 187
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how u got c 52
It is a mathematical process. It is the concept of modular arithmetic in mathematics.
Hi may I know why you choose p=13 as e while not the other number as both p and q, their GCD with phi 143 are the same, which is 1
You can take any number...This is just example....
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Thank you
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@@ChiragBhalodia I subscribed you my lord
Thank you so much my dear friend..
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Hi sir,
If P and Q are not given shall we consider any prime number by own?
Is plain text is same as message m?
Kindly reply I have an exam
sorry for late reply... you can assume the prime number...
@@ChiragBhalodia plain text and message m are same sir?
Yes it is same.... Also it may be given in binary format.... if plain text in binary then first you must have to convert in decimal.
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sir plz tell how to find value of 13^13mod143
Hello Anand, I have solved using modular arithmetic method....
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plain text kaise find karenge
Formula is already given in algorithm...
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C=52 how to we get
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How to find the mod of those numbers ??
I have find this mod using modular arithmetic theorem...
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Any imps for INS tomorrow ?
Most IMP for tomorrow examination : All cipher techniques,active and passive attack, des, aes, diffie Hellman, rsa, application of digital signature, kerberos, DSA, public key distribution, hmac, all block cipher modes, ssh, SSL
@@ChiragBhalodia thanks!
www.edtechnology.in/data-mining-gtu-study-material/
For data mining important
@@ChiragBhalodia wow thanks for remembering. Going through it now!
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c=41
que..1
True.....This video share with others....Subscribe my channel..
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plain text ki value kaha se laye sir aap ?
In which example?
@@ChiragBhalodia 5:07 sir how to find plain text
In examination Plain text will be given in question....refer questions given in videos......here I am explaining example...So, I have taken plain text = 13.
Most IMP for tomorrow examination : All cipher techniques,active and passive attack, des, aes, diffie Hellman, rsa, application of digital signature, kerberos, DSA, public key distribution, hmac, all block cipher modes, ssh, SSL
@@ChiragBhalodia are sir !!!!!!!!!!❤️ ❤️ ❤️ ❤️ ❤️ ❤️ ❤️ thanks a lot
Sir how to find the modulus
Modulus is find using arithmetic modular method...... Please share this video with your friends...
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Sorry professor why is "e" in excercise2 the encryption key why didn't say it's the public key directly?
Dear both are same.....
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Sir if you having ppt please send it
Contents available on my blog...
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Step 1 ma q ni value 11 hati to 3rd step ma 12-1 kemnu aave @Chirag Bhalodiya?
Dear it was my typing mistake.... See the further proces.. It has written 10....so it is 11-10....
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can someone pls explain why we stopped searching for d when we reached 37??
As per the algorithm, Value of d must be prime and integer number.. So, I stopped finding d=37.
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