Modular exponentiation

Using the repeated squaring algorithm to calculate 2^300 mod 50.

[@VFXCommander]

Perfect explanation, despite the twitching.

[@richardblack1588]

Diego de la Vega flickering

[@Peachyj06]

spasming

[@Pixelynx]

orgasiming Wait... What were we doing?

[@SurajKumar-bw9oi]

glitching

[@arampathkushan2861]

tweaking

[@fzndn-xvii]

When a KZbin video explains explains better than your professor. Thanks!

[@outlaws9295]

Besides the occasional issues with the video itself, this was a very great demonstration. Excellent job!

[@rcollins0618]

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).

[@outlaws9295]

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.

[@FRANCISCORODRIGUEZ-vq2jj]

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!

[@ZacMitton]

Or 5 minutes at 2x speed

[@ianchen1962]

@@ZacMitton haha

[@ko-Daegu]

@@ZacMitton I watched at 1.5 while skipping

[@MyOneFiftiethOfADollar]

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.

[@alexisreynolds4277]

Finally a good example! I looked at so many videos that didn't help before finding this one. Thank you!

[@terriblast8076]

One of the best tutorial I've watch..You discuss well

[@sampugh6404]

This just saved me on security homework. Thank you so much. Very cool concept

[@joshuagroeschl7001]

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.

[@deano50z]

Excellent video, followed along with a notepad and pen and understood it less than 10 mins later. Thanks pal!!

[@amarthya.t9842]

Even after 9 years of upload this is just great demonstration. thank you mate.

[@yaggikjalan]

This helped me so much for rsa algorithm questions! Thanks a million ☺

[@jsteezeful]

This is so good it also makes me question my professor and their pathetic book that tries to explain this in 2 paragraphs.

[@simonliu2268]

Amazing thank you very much, I didn't understand my professors abbreviation of this, but you did a very VERY good job.

[@NOSHEDMANTIS]

Great Video, clear explanation and good audio quality (essential).

[@alexrobert4614]

Best tutorial out there. My prof sucks. Took me 2 days to understand fully. Thank you

[@psychedelicfungi]

One very helpful video! Thanks! (Unfortunately the screen flicker was really distracting)

[@tonychen1083]

Thank you for your excellent explanation!

[@shinwoo_park]

it helps me a lot! thank you sir. What a nice lecture

[@stea27]

Very informative, easy to understand it in your demonstration. Thank you!

[@monowar6723]

Thanks a lot ! i will remember this video for the rest of my life.

[@JoseSanchez-vv1zd]

Great job! Thanks for the clear explanation! :)

[@user-vq1ef8kn9m]

That was an awesome explanation. Solved my problem~.

[@memento_mori6019]

Excellent breakdown. Thank you.

[@SpookyRumi]

Amazing! You saved me so much time with this

[@kdpoint4221]

it took 3h to get a perfect explanation..... thanks a lot

[@ianchen1962]

dang that's so cool. Learned a lot from this

[@sebastianrauch2758]

Thanks for the good explanation. Realy helped a lot. :D

[@rafsanjanimuhammod309]

Really Nice One, Sir !!!

[@aries3690]

Thank you so much for your clear explanation!

[@Th1sUsernameIsNotTaken]

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.

[@harley3514]

Thank you, other videos on this were not making sense for me, but this did

[@EndietheEnderman]

Helped heaps! Thanks

[@jonty3551]

Great explanation man..... appreciate it🤝👍

[@stephaniagambaroff820]

excellent explanation! Thank you.

[@priyankmungra29]

nice explanation. Thank you for your efforts.

[@costerchimbwanda3936]

saw the screen flickering..thought t was a problem with my pc..anyway was a great illustration

[@hardinrami8500]

Thank you for the explanation! It took me a while to get it though.

[@DVRhm25]

I watched this drunk. It all makes sense. Nice work prof!

[@shailendrakumar3130]

Thanks man, searching for this information from mornimg

[@anhminhtran3438]

Very well explained. Thank you

[@krgaming1650]

Thank you, you saved my grade on today's exam.

[@rajatsingh8143]

best explanatioon in the world saved me for exam

[@SitWithItBob]

This helped a lot, thanks!

[@b34amanahmed]

He is brilliant and excellent Once look into this u will find ur solutions

[@KoksMOks]

that alg is amazing AF

[@RobinDuda]

Beautiful.

[@ericstroud1]

Great video!

[@navyakakarlapudi6126]

Nice explanation sir.Thank u so much......

[@owaisali995]

really thankfull for this video.. great explanantion...

[@tinkerbajwa]

wow I understood it right away! Thankyou so much...

[@subizsunuwar7068]

thanks sir ....it's help me, while calculating encryption of msg.

[@dialamark7863]

I loved it. Thank you

[@matkenis3013]

I was looking for modular exponentiation explanation all over youtube and it all pretty much was garbage. Thank you for actually explaining things ffs :)

[@Chriib]

Cool trick. I will use this trick on my exam today.

[@Just_an_alien1]

Helped me a lot thank you

[@ninjanothing8343]

Thank you very much! Very helpful

[@meysamamini9473]

Was very vary useful Thanks a Lot !

[@AlexandrBorschchev]

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.

[@AlexandrBorschchev]

haha, cool story buddy.

[@KhandkarAsifHossain]

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.

[@vineet1034]

what happen if base is 132 or bigger

[@ketanbhalerao1795]

Damn Good ! Thanks a Bunch, man !!

[@EasiestTutor]

thank you rhabk you i was struggling this was best and easiest

[@Sahandkeshavarz]

how would i do that if my exponent is greater than 255 then i wont be able to convert to binary

[@lilbpack2100]

brilliant. thank you!!

[@MichelleVo]

How did you figure out the binary for 200 from scratch?

[@hungryghost5589]

great video, thanks

[@nikhilchoudhary277]

WOOOOOWWWWW! What an Explainantion

[@kriswalsh1020]

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.

[@mdmuktadirmazumder284]

love you 3000mod3000

[@jiaxinli1674]

Thanks for saving my ICS 6D

[@syednawaid6980]

nice explanation.....

[@urensoft6662]

you are just awesome...

[@ramkumarnj7617]

Excellent!

[@swordclash6618]

The explanation is A+ but the screen flickering almost made me go nuts.

[@directx872]

This is so fucking awesome. I now have *the power to compute*

[@zanonymousruiz9697]

This saved my entire ass- Thanks!

[@beasty45]

Thank you , Thank you, Thank you!!!!

[@iuseyoutubealot]

u are a legend

[@cbabrunal]

In a few minutes you explain what my university professor couldn't in months

[@TariqAbdulGhani-kt1nk]

very helpful thank u🙂

[@ayaanpuri5697]

lots of really large exponents' modulos are equal to 1 by fermat's little theorem

[@DanielVazquez]

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.

[@tanishqraina3656]

Well go ahead and use binomial theorem

[@zaferkhourdaji1958]

Best Explanation

[@salounik.2894]

AMAZING!!!!!!

[@SHIVASURYA]

really useful one :)

[@maidul13]

Thank you!

[@Mohammed-hw4kz]

I love you, thanks! That 6 marks :)

[@rishabhbharti7491]

Crystal clear.

[@larsmees4952]

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?

[@abhigyanraha5620]

wow! amazing.

[@mohammadahmedragab837]

great explanation. please, in the last part of the video, Why 3751 mode 50? Should 3751 the answer of 3^200 mode 50 ? I need to understand this more

[@MotorHeadTech]

I don't get the 31 and 11 parts I get everything else tho please help

[@dominikchynoradsky4228]

amazing, thanks

[@stanger8992]

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?

[@cameronsaunders9020]

The flicking is really bothering my eyes lol

[@ScotMatson]

This video is so thorough and one of the better ones I have come across but that screen flicker practically ruins it for me, it is really dizzying by the end. Such a shame! Thanks for taking the time to put this together otherwise, my book cannot nearly present this concept in such a clean fashion as you have here.

[@skurt21]

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

[@skurt21]

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)

[@jeffryyapin1399]

One Question How about if the exponent was smaller then 128. for example 120 did we need to know the 128bit??

[@sgut1947]

+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.

[@diegoi1313]

Thanks!