Such a simple demonstration really beats trying to learn this from just reading formal notation. thank you very much! (I often have this trouble with discrete math - it's not hard stuff, I just get caught up in keeping all the variables in my head).
Exact same issue here, man. I took one look at this process (or at least a very similar one) in my textbook and it made little to no sense. Once I saw this video, the process became crystal clear. I know that the formal notation is mathematically correct, but it's usually not the best way to demonstrate a concept for the first time.
This is most stupendous indeed, I was struggling with this conceptually for some time before stumbling across this video. I appreciate you for taking the time to curate these works! Excellent example and explanation.
The difference is you did not sleep during this video AND of course it was your professor's fault for not being able to keep you awake. Be accountable for your own learning outcomes rather than blaming others.
Thank you for this! other video's I've seen just completely skip steps or don't explain. Can't get a meeting with my teacher for a few days (online learning) and the book didn't explain any of the random jumping it was doing (didn't do steps, just jumped to the "solved" part). Now I can actually practice lol.
I was looking for modular exponentiation explanation all over youtube and it all pretty much was garbage. Thank you for actually explaining things ffs :)
Ok that is kinda obvious but i have an exam soon where i will have to calculate 30 of such numbers without using calculator, and it has to take max 10 min because it's one of 18 excersises on that exam. How do i go about solving for example 33^350 mod 7 in 20 seconds, using only pen and paper?
Wouldn't it be faster to first perform a modular division using the same value (50) on the exponent? 2^(200 mod 50) = 2^0 = 1 Or is this just a coincidence?
I found a problem, and since nobody mentioned it in 9 year, it's probably on my side. I follow the remainders or moduli (?sorry) for powers up to 32. But according to my calculations 5 to the power of 64 mod 50 is 20 and not 31 as you mention. Am I really wrong? I calculated it in powershell
+Jeffry Yapin No, you don't need the 128, because 120 = 64+32+16+8 (binary 1111000). In general, you only need up to the power of two that's less than the required exponent.
Perfect explanation, despite the twitching.
Diego de la Vega flickering
spasming
orgasiming
Wait... What were we doing?
glitching
tweaking
When a CZcams video explains explains better than your professor. Thanks!
Besides the occasional issues with the video itself, this was a very great demonstration. Excellent job!
Such a simple demonstration really beats trying to learn this from just reading formal notation. thank you very much! (I often have this trouble with discrete math - it's not hard stuff, I just get caught up in keeping all the variables in my head).
Exact same issue here, man. I took one look at this process (or at least a very similar one) in my textbook and it made little to no sense. Once I saw this video, the process became crystal clear. I know that the formal notation is mathematically correct, but it's usually not the best way to demonstrate a concept for the first time.
This is so good it also makes me question my professor and their pathetic book that tries to explain this in 2 paragraphs.
Finally a good example! I looked at so many videos that didn't help before finding this one. Thank you!
Even after 9 years of upload this is just great demonstration. thank you mate.
This is most stupendous indeed, I was struggling with this conceptually for some time before stumbling across this video. I appreciate you for taking the time to curate these works! Excellent example and explanation.
my professor wasn't able to teach me in a 3 hour class period what you just taught me in 11 min and 36 seconds. thank you so much!
Or 5 minutes at 2x speed
@@ZacMitton haha
@@ZacMitton I watched at 1.5 while skipping
The difference is you did not sleep during this video AND of course it was your professor's fault for not being able to keep you awake.
Be accountable for your own learning outcomes rather than blaming others.
Excellent video, followed along with a notepad and pen and understood it less than 10 mins later. Thanks pal!!
One of the best tutorial I've watch..You discuss well
This just saved me on security homework. Thank you so much. Very cool concept
One very helpful video! Thanks!
(Unfortunately the screen flicker was really distracting)
Best tutorial out there. My prof sucks. Took me 2 days to understand fully. Thank you
This helped me so much for rsa algorithm questions!
Thanks a million ☺
Amazing thank you very much, I didn't understand my professors abbreviation of this, but you did a very VERY good job.
Great Video, clear explanation and good audio quality (essential).
Thank you for your excellent explanation!
Thanks a lot ! i will remember this video for the rest of my life.
it helps me a lot! thank you sir. What a nice lecture
Excellent breakdown. Thank you.
Amazing! You saved me so much time with this
Great job! Thanks for the clear explanation! :)
Very informative, easy to understand it in your demonstration. Thank you!
it took 3h to get a perfect explanation..... thanks a lot
Thank you, other videos on this were not making sense for me, but this did
That was an awesome explanation. Solved my problem~.
Thanks for the good explanation. Realy helped a lot. :D
dang that's so cool. Learned a lot from this
Thank you so much for your clear explanation!
Helped heaps! Thanks
saw the screen flickering..thought t was a problem with my pc..anyway was a great illustration
Thank you for this! other video's I've seen just completely skip steps or don't explain. Can't get a meeting with my teacher for a few days (online learning) and the book didn't explain any of the random jumping it was doing (didn't do steps, just jumped to the "solved" part). Now I can actually practice lol.
Beautiful.
Thank you for the explanation! It took me a while to get it though.
I watched this drunk. It all makes sense. Nice work prof!
what happen if base is 132 or bigger
excellent explanation! Thank you.
Awesome video. But I am having trouble finding the mod with high base number such as
26^37 mod 77
can u help with that.
26^8 seems to pose a problem.
Thank you, you saved my grade on today's exam.
Very well explained. Thank you
nice explanation. Thank you for your efforts.
On the step where you do 200+128+64+8 how'd you get the 64 and 8. I have a question similar but it's 221.
This helped a lot, thanks!
Great explanation man..... appreciate it🤝👍
how would i do that if my exponent is greater than 255 then i wont be able to convert to binary
He is brilliant and excellent
Once look into this u will find ur solutions
Really Nice One, Sir !!!
How did you figure out the binary for 200 from scratch?
that alg is amazing AF
Thanks man, searching for this information from mornimg
best explanatioon in the world
saved me for exam
I loved it. Thank you
wow I understood it right away!
Thankyou so much...
Great video!
I was looking for modular exponentiation explanation all over youtube and it all pretty much was garbage. Thank you for actually explaining things ffs :)
Cool trick. I will use this trick on my exam today.
love you 3000mod3000
Thank you very much! Very helpful
really thankfull for this video.. great explanantion...
Nice explanation sir.Thank u so much......
Thanks for saving my ICS 6D
Thank you!
Ok that is kinda obvious but i have an exam soon where i will have to calculate 30 of such numbers without using calculator, and it has to take max 10 min because it's one of 18 excersises on that exam. How do i go about solving for example 33^350 mod 7 in 20 seconds, using only pen and paper?
thanks sir ....it's help me, while calculating encryption of msg.
Helped me a lot thank you
Was very vary useful Thanks a Lot !
This is so fucking awesome. I now have *the power to compute*
Excellent!
brilliant. thank you!!
The explanation is A+ but the screen flickering almost made me go nuts.
Damn Good ! Thanks a Bunch, man !!
Is there a shorter way? I mean, I just had a test and I was expected to solve 2^75 (mod 73) in one minute or so.
Well go ahead and use binomial theorem
WOOOOOWWWWW! What an Explainantion
thank you rhabk you i was struggling this was best and easiest
Wouldn't it be faster to first perform a modular division using the same value (50) on the exponent? 2^(200 mod 50) = 2^0 = 1
Or is this just a coincidence?
great video, thanks
This saved my entire ass- Thanks!
u are a legend
Crystal clear.
you are just awesome...
Thank you , Thank you, Thank you!!!!
very helpful thank u🙂
Best Explanation
How did you get that binary number? 11001000?
nice explanation.....
I don't get the 31 and 11 parts I get everything else tho please help
Thanks!
wow! amazing.
lots of really large exponents' modulos are equal to 1 by fermat's little theorem
AMAZING!!!!!!
Im grade 9 learning this for a math contest. It looks crazy useful and really really cool, i want to know how to master this. Thank you for the video.
haha, cool story buddy.
amazing, thanks
I don't understan why you replaced the 3^8 with 31, why are they the same?
In a few minutes you explain what my university professor couldn't in months
really useful one :)
I found a problem, and since nobody mentioned it in 9 year, it's probably on my side.
I follow the remainders or moduli (?sorry) for powers up to 32.
But according to my calculations 5 to the power of 64 mod 50 is 20 and not 31 as you mention.
Am I really wrong?
I calculated it in powershell
yep, I'm the problem, sorry.
I have no idea what I did wrong, but I'm getting the same numbers as you do now. Sorry (y)
Awesome
One Question
How about if the exponent was smaller then 128. for example 120 did we need to know the 128bit??
+Jeffry Yapin No, you don't need the 128, because 120 = 64+32+16+8 (binary 1111000). In general, you only need up to the power of two that's less than the required exponent.
Good video. TY!