Modular exponentiation

[@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

[@outlaws9295]

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

[@fzndn-xvii]

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

[@joshii4537]

Lmao I had a course where the prof just showed us this guys videos

[@amarthya.t9842]

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

[@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.

[@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.

[@jsteezeful]

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

[@sampugh6404]

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

[@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)

[@alexisreynolds4277]

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

[@deano50z]

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

[@kdpoint4221]

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

[@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.

[@terriblast8076]

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

[@NicholasGorsich]

from 2024, incredible video, helped me out lots!!!

[@rajatsingh8143]

best explanatioon in the world saved me for exam

[@DVRhm25]

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

[@b34amanahmed]

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

[@krichard12]

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

[@costerchimbwanda3936]

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

[@jiaxinli1674]

Thanks for saving my ICS 6D

[@NOSHEDMANTIS]

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

[@cbabrunal]

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

[@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 :)

[@SpookyRumi]

Amazing! You saved me so much time with this

[@simonliu2268]

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

[@KoksMOks]

that alg is amazing AF

[@sebastianrauch2758]

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

[@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?

[@yaggikjalan]

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

[@monowar6723]

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

[@Chriib]

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

[@directx872]

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

[@swordclash6618]

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

[@cameronsaunders9020]

The flicking is really bothering my eyes lol

[@hardinrami8500]

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

[@priyankmungra29]

nice explanation. Thank you for your efforts.

[@ayaanpuri5697]

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

[@vineet1034]

what happen if base is 132 or bigger

[@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.

[@shinwoo_park]

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

[@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

[@jonty3551]

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

[@victorakinyemi2692]

To calculate 2^300 mod 50. easy approach to solve this question is to Fermat thereom: a^ phi(n) = 1 mod n with gcd(a,n)= 1 so since gcd(2,50) = 1 then i calculate phi(50)= (2-1)*(5^2-5) =20 so 2^20 =1 mod 50 if i multiply this modular expression 15 times then i have 2^300 = 1 mod 50

[@killator3421]

One question how can 2 multiplied several times with 2 be an odd number when the mod is 50?

[@johnchang8279]

This explanation is a little confused. First, the problem is 3^200, not 2^300. But if you want to find 2^300 (mod 50), gcd(2,50)=2, so you can't use Euler directly. Instead you have to factor the modulus into the powers of its primes: x ≡ 2^300 (mod 25) and x ≡ 2^300 (mod 2). Then apply Euler to the first congruence: x ≡ 1 (mod 25). The second congruence is x ≡ 0 (mod 2). The solution is x ≡ 26 (mod 50).

[@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.

[@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.

[@harley3514]

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

[@robertniyazoff2591]

Why did you convert 200 to binary? Is it necessary to do? You didn't use it in the problem to help you

[@tonychen1083]

Thank you for your excellent explanation!

[@navyakakarlapudi6126]

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

[@stea27]

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

[@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?

[@sindisiwemncube4665]

How did you get that binary number? 11001000?

[@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.

[@memento_mori6019]

Excellent breakdown. Thank you.

[@owaisali995]

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

[@吳振瑋-y4l]

That was an awesome explanation. Solved my problem~.

[@mdmuktadirmazumder284]

love you 3000mod3000

[@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)

[@shailendrakumar3130]

Thanks man, searching for this information from mornimg

[@tinkerbajwa]

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

[@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.

[@aries3690]

Thank you so much for your clear explanation!

[@ianchen1962]

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

[@Sahandkeshavarz]

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

[@figurativelyrazzmatazz7656]

I don't understan why you replaced the 3^8 with 31, why are they the same?

[@anhminhtran3438]

Very well explained. Thank you

[@masutthe_gooner3356]

What if the power is too long what then?

[@wideputin6586]

What if the power is a big prime number

[@kelvinella]

this can be done in one line... Since (3, 50) = 1, and phi(50) = 50*1/2*4/5 = 20, then by Euler's thm, 3^200 = 1 mod 50

[@JoseSanchez-vv1zd]

Great job! Thanks for the clear explanation! :)

[@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

[@stephaniagambaroff820]

excellent explanation! Thank you.

[@MotorHeadTech]

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

[@zanonymousruiz9697]

This saved my entire ass- Thanks!

[@MichelleVo]

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

[@beasty45]

One Question. In the Modular exponentiation video, what if the exponent was an odd number such as 201 instead of 200. Is the procedure of solving the problem the same?

[@Jambo601165]

It would be, you would just have to do the calculation for 3^1 mod 50 = 3 mod 50. Then at the end when you are multiplying your exponents you would add 3^1 to the calculation, so 3^128 x 3^64 x 3^8 x 3^1 = (11)(31)(11)(3) = 11253 mod 50 = 3

[@ouraghyoussef5612]

+Jamie Watkinson 128+64+8+1=201!! ???

[@AkshayAradhya]

Would have liked the video if it wasnt for the flickering. Please use a better screen cast cause you teach really well and that shouldnt be spoiled by some flickering screen.

[@rafsanjanimuhammod309]

Really Nice One, Sir !!!

[@meysamamini9473]

Was very vary useful Thanks a Lot !

[@SitWithItBob]

This helped a lot, thanks!

[@ai_robozen]

Beautiful.

[@KayOScode]

You still end up needing a big integer-like library when it comes to RSA because your working with bases and powers hundreds or thousands of bits long

[@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.

[@EasiestTutor]

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

[@EndietheEnderman]

Helped heaps! Thanks

[@subizsunuwar7068]

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

[@jeanantazo7746]

what happen in 3751?

[@ninjanothing8343]

Thank you very much! Very helpful

[@Just_an_alien1]

Helped me a lot thank you

[@zaferkhourdaji1958]

Best Explanation

[@manny1tm]

What if the exponent is a prime? 417^103 mod 779.

[@jesuschristsonofgod9709]

+Manuel Martin Try Fermat's Little Theorem

[@anujaghosh179]

can you show it here using fermat's?

[@eodkybb]

hei

[@johnchang8279]

Factor the modulus 779 = 19*41, use modularity to reduce it to 417^103 ≡ (-1)^103 ≡ -1(mod 19) and Fermat's little theorem on 417^103 ≡ 417^23 ≡ 7^23 ≡ 7*49^11 ≡ 7*8^11 ≡ 7*2^33 ≡ 7* 2^(10*3) * 8 ≡ -56 (mod 41). Then solve the simultaneous congruences: -56+41k ≡ -1(mod 19). 41 ≡ 3 (mod 19) and 55 ≡ -2 (mod 19), so this reduces to 3k ≡ -2 (mod 19). 3k ≡ -2+2*19 ≡ 36 (mod 19). Dividing by 3 gives k≡12 (mod 19), so x≡ -56+41*12 ≡ 436 (mod 779). By contrast, exponentiation by repeated squaring would be far more laborious, suitable for a computer but quite tedious for a human. On the other hand, repeated squaring is the method to use if the modulus is a large prime, larger than the exponent. Euler's theorem makes the stated problem x≡ 3^200 (mod 50) almost trivial. It says 3^20 ≡ 1 (mod 50), because the totient of 50= 50(1-1/2)(1-1/5)=20 and gcd(3,50)=1.

[@dialamark7863]

I loved it. Thank you

[@nikhilchoudhary277]

WOOOOOWWWWW! What an Explainantion

[@pc557]

You should put a photo-epilepsy warning on this video!

[@abaleg2272]

it took this guy 11 minutes to do something i did in 30 seconds. 3^5=43=-7(mod 50), and squaring both sides we get 3^10= 49=-1(mod 50). taking the 20th power on both sides, we get (3^10)^20=(-1)^20(mod 50) Thus, 3^200= 1(mod 50) And i'm in grade 9 learning this for the IMO

[@INAYATULLAHSHEIKH]

Shortcut: 3^200 is cong to 7^40 is cong to 1 (mod 50).So remainder =1.

[@rishabhbharti7491]

Crystal clear.

[@TariqAbdulGhani-kt1nk]

very helpful thank u🙂