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.
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.
I was looking for modular exponentiation explanation all over youtube and it all pretty much was garbage. Thank you for actually explaining things ffs :)
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.
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
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).
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.
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
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?
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.
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?
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
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?
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
Perfect explanation, despite the twitching.
Diego de la Vega flickering
spasming
orgasiming
Wait... What were we doing?
glitching
tweaking
Besides the occasional issues with the video itself, this was a very great demonstration. Excellent job!
When a TH-cam video explains explains better than your professor. Thanks!
Lmao I had a course where the prof just showed us this guys videos
Even after 9 years of upload this is just great demonstration. thank you mate.
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.
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!
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.
Best tutorial out there. My prof sucks. Took me 2 days to understand fully. Thank you
This just saved me on security homework. Thank you so much. Very cool concept
Excellent video, followed along with a notepad and pen and understood it less than 10 mins later. Thanks pal!!
from 2024, incredible video, helped me out lots!!!
One of the best tutorial I've watch..You discuss well
One very helpful video! Thanks!
(Unfortunately the screen flicker was really distracting)
This is so fucking awesome. I now have *the power to compute*
it took 3h to get a perfect explanation..... thanks a lot
Amazing thank you very much, I didn't understand my professors abbreviation of this, but you did a very VERY good job.
The flicking is really bothering my eyes 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 :)
This helped me so much for rsa algorithm questions!
Thanks a million ☺
Amazing! You saved me so much time with this
I watched this drunk. It all makes sense. Nice work prof!
Thanks a lot ! i will remember this video for the rest of my life.
Great Video, clear explanation and good audio quality (essential).
In a few minutes you explain what my university professor couldn't in months
saw the screen flickering..thought t was a problem with my pc..anyway was a great illustration
Thanks for the good explanation. Realy helped a lot. :D
The explanation is A+ but the screen flickering almost made me go nuts.
Very informative, easy to understand it in your demonstration. Thank you!
best explanatioon in the world
saved me for exam
He is brilliant and excellent
Once look into this u will find ur solutions
that alg is amazing AF
Thank you, other videos on this were not making sense for me, but this did
Thank you for the explanation! It took me a while to get it though.
Thank you, you saved my grade on today's exam.
Thanks for saving my ICS 6D
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.
wow I understood it right away!
Thankyou so much...
lots of really large exponents' modulos are equal to 1 by fermat's little theorem
Thank you for your excellent explanation!
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.
That was an awesome explanation. Solved my problem~.
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.
Excellent breakdown. Thank you.
dang that's so cool. Learned a lot from this
Thank you so much for your clear explanation!
it helps me a lot! thank you sir. What a nice lecture
nice explanation. Thank you for your efforts.
Great explanation man..... appreciate it🤝👍
Great job! Thanks for the clear explanation! :)
Cool trick. I will use this trick on my exam today.
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
One question how can 2 multiplied several times with 2 be an odd number when the mod is 50?
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).
Thanks man, searching for this information from mornimg
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.
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
Nice explanation sir.Thank u so much......
really thankfull for this video.. great explanantion...
what happen if base is 132 or bigger
Very well explained. Thank you
excellent explanation! Thank you.
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
This saved my entire ass- Thanks!
thank you rhabk you i was struggling this was best and easiest
This helped a lot, thanks!
love you 3000mod3000
Was very vary useful Thanks a Lot !
Warning , This video has been identified to potentially trigger seizures for people with photosensitive epilepsy.
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
Helped heaps! Thanks
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?
Really Nice One, Sir !!!
Shortcut: 3^200 is cong to 7^40 is cong to 1 (mod 50).So remainder =1.
I loved it. Thank you
Thank you very much! Very helpful
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.
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
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.
Beautiful.
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
thanks sir ....it's help me, while calculating encryption of msg.
Fuck, this was beautiful... I have to say thank you!
Great video!
You should put a photo-epilepsy warning on this video!
Helped me a lot thank you
Damn Good ! Thanks a Bunch, man !!
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?
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)
WOOOOOWWWWW! What an Explainantion
great video, thanks
how would i do that if my exponent is greater than 255 then i wont be able to convert to binary
Best Explanation
I don't get the 31 and 11 parts I get everything else tho please help
very helpful thank u🙂
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?
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
+Jamie Watkinson
128+64+8+1=201!! ???
Excellent!
brilliant. thank you!!
nice explanation.....
you are just awesome...
What if the power is a big prime number