Public Keys Part 2 - RSA Encryption and Decryptions

Simple, clear explanation. Thank you.

i dont understand how she got the common values from 1=5-2(7)+2(5) to 1=3(5)-2(7) ???

Why is e = 7?

How do you calculate 28^23 using calculate to get decimal. Im getting exponential figures

you just misplaced the numerical represetations of c and d.

You gave me exactly what I was looking for. Thank you so much. Hope you have an amazing day <3

So why will 3rd party person not be able to decrypt like Alice???

great ! good job

How is H 7 ...or are we saying A is 0 ...which is a bit confusing

she sounds like the queen

D:

28^23%55=23 not 7 dude what's up with her......


or am I calculating 28^23%55 wrong, b'coz i don't know how to calculate % sums

understandable

How did you calculate -17mod40?

Great video, thanks a lot

it's helpful , thank's a lot but why our options are 3,7,11,13,17 we have also 19,23,......

(34^2-1)/55=21. (21^2-1)/55=8. 8*21+1=13^2. 55=21+(2*13)+8 and 34+21=55 . Gcd(55,33)=11 and gcd(55,20)=5

28^23 =7 hows dat

hi, you have some severe error there.. (M^7|55)^23|55 == (M^7|55)^3|55 == (M^7|55)^43|55

so 7^7|55 = 28 ... 28^3|55 = 7

i think that the private key should be better secured then this..

Thanks a lot, it's so much helpful.......

how is 7^7mod55=28mod55?

why is 8^7 mod 55 = 2 mod 55? explain to me plz!

That Extended Euclidean Algorithm is pain in the brain.

i dont get it how you get -17 to 23 mod 40

In the options for 'E' why is 9 there ?

Thanks a zillion madam. This is great.

The teacher made a subtle technical error. If H=7 I=8 D=3 and E=4 then A must be 0. But 0^7 = 0 mod 55 = 0. If, in such a basic code system, you saw 0's you'd be able to tell straight away that they are As, and a good codebreaker would be able to figure everything out. The alphabet needs to be 1-based, not 0-based.

i dont get how to do decryption when its numbers like 49^23 etc i cant get the remainder on calculator to get 4 mod 55 please help

Thanks! This video helped alot!

how the hell did you get 28?? as C