This was honestly so helpful. I could not for the life of me understand how modulo notation worked, and seeing this explained with the theorem got me unstuck in the middle of a project. Thank you so much!
Sometimes it’s better to see an application first, and only thereafter lean into the theory. Thanks, Jay, for starting me on the path towards understanding this beautiful theorem. Your examples were incredibly poignant 🏆🙌🏽🎊
Thank you so much! I have an exam tomorrow and homework due the next day and you just saved me! You explained so good! I worked your examples along with you...
Thank you so much, I am a native spanish speaker so I had been looking for spanish videos to figure that out and none of those videos worked for me and I found your video and it really helped me out!! . Finally I just got it
i am from india jay, a student. We learn this type of things in senior High School (11th standard) but the methods are harder(Binomial theorem). This method is great.
Maths with Jay in countries such as India complex topics like number theory are introduced to students as early as secondary school. But these are just introducing the topic. When we get to college/uni that is when we learn it at the advanced level.
Have i understood correctly here: 7^50 (mod 13) Since 7^12 = 1 mod 13 50 = 4 * 12 + 2 So 7^50 = 7^(4*12+2) = 1^4 * 7^2 = 1 * 49 = 10 mod 13 So least residue is 10 ?
THANK YOU THANK YOU THANK YOU! I'm in my final year as an undergrad in Computer Science and have been struggling dreadfully with congruences and residues. YOU ARE A LIFESAVER!
thankyou , i could not understand this leeson throughout my lectures in uni, then i looked into your video and i'm now so capable of doing this kinda sums.
Coud you please explain why remainder does not stay the same for all primes? I tried your example with 2^50 breaking it down with all primes from 17 to 3. For all except 7 and 3 I got remainder 4, which means its composite. But for 7 and 3 I got remainder 1, which means its Prime. I will show it on paper: 2^50 = 2^(10*5) = 1 for P = 11 2^50 = 2^(2*25) = 1 for P = 3 Is there some kinda of rule for choosing primes you are checking for?
Maths with Jay I've been looking around for something particular, maybe you know the answer. How can you do this exact thing but without prime numbers? For example, Find the remainder of 9^342 | 10
+Jake Moore This is a different situation, so think about it in a different way: Start by writing down the first few powers of 9, find the remainders when you divide by 10, and you will see a simple pattern...
The point is that we are considering whether or not the answer can be displayed on a scientific calculator....we start by looking at numbers where the remainder is shown on a calculator, then we look at an example where the numbers can be input without generating an error message, then we look at an example where the calculator cannot evaluate...we are not trying to look at the largest possible example!
Hi! i have a problem and Im not sure whether its related to this or not. if M^e mod N, N is not a prime number and extremely large, but not larger than e. M is a four digit number, and im trying to so M^e mod N, can I still use this? will the remainder pattern work?
You are a fantastic teacher! Thanks so much.
Thank you very much for your fabulous feedback.
This was honestly so helpful. I could not for the life of me understand how modulo notation worked, and seeing this explained with the theorem got me unstuck in the middle of a project. Thank you so much!
Excellent! Thank you for letting us know - it is much appreciated.
The way you explain is nothing short of brilliant.
Wow! Thank you T J
Sometimes it’s better to see an application first, and only thereafter lean into the theory. Thanks, Jay, for starting me on the path towards understanding this beautiful theorem. Your examples were incredibly poignant 🏆🙌🏽🎊
Glad to help! And thanks a lot for your fantastic feedback
I am a French student and this was the most helpful video I found, your explanation was clear and simple, I finally understood. Thank you !
@Yanna Laurent: Thank you so much...merci beaucoup!
i ve seen 100 of videos and website this is by far the easiest solution
@KAK: Thank you for your great feedback!!
Thank you so much! I have an exam tomorrow and homework due the next day and you just saved me! You explained so good! I worked your examples along with you...
Brilliant! Good Luck for your exam, Carlos
Thanks Jay one of the best explanation I was in youtube you made it look very easy thanks...
Great to hear!
You have saved my life! I needed to know how to do this for my Mathcounts, and I didn't know! Thank you so much!
Wow...great to know we're saving lives!
:D
You have saved me so much time and effort preparing for a test; thank you so much!
You are so welcome!
I've been struggling with this topic for days now in preparing to lecture this topic.... you've made it so easy to understand. Keep up the good work!
Glad it was helpful!
Now with SUBTITLES
Thank you so much, I am a native spanish speaker so I had been looking for spanish videos to figure that out and none of those videos worked for me and I found your video and it really helped me out!! . Finally I just got it
Glad I could help! ¡Estoy aprendiendo español!
Again, Maths with Jay saves the day. You are such a huge help, thank you 😊
You are so welcome!
Mind blowing explanation.
Really
Thank you so much 😀
Loved it. Now I know the application of this theorem.
@Kunal Singh Gusain: Excellent!
I love your Voice and really you are fantastic teacher I had ever met in my life
Thanks for the video
@Rishi: Thank you!
Thanks a ton, my discrete math exam is in 2 hours and this was the last topic that was really giving me trouble and you cleared it right up
@Bobboy6: Excellent - good luck with the exam!
i am from india jay, a student. We learn this type of things in senior High School (11th standard) but the methods are harder(Binomial theorem). This method is great.
@Divyanshu: That's interesting. In England this sort of work is not done at school, only at university.
Maths with Jay in countries such as India complex topics like number theory are introduced to students as early as secondary school. But these are just introducing the topic. When we get to college/uni that is when we learn it at the advanced level.
@FAKEOUT DRAGON: Thank you for explaining that.
can you tell that method to me?
Thank you very much for such a clear and an elegant explanation!
@altair ezio: Thank you!
This was extremely helpful having three levels of examples!
Great to know!
Thank you!! this really helped for my discrete mathematics class!
That's great! Thank you so much for letting us know.
nice
Thank you very much, helped me a lot for my first year of computer science study in france !
Best video on the subject on the 4 i watched
Great to hear!
Have i understood correctly here:
7^50 (mod 13)
Since 7^12 = 1 mod 13
50 = 4 * 12 + 2
So 7^50 = 7^(4*12+2)
= 1^4 * 7^2
= 1 * 49
= 10 mod 13
So least residue is 10 ?
Thanks for making this video. This is well explained. You are a great teacher!
Thank you so much ♥️
THANK YOU THANK YOU THANK YOU! I'm in my final year as an undergrad in Computer Science and have been struggling dreadfully with congruences and residues. YOU ARE A LIFESAVER!
N3rdRag3: Thank you! Good luck with the rest of your studies
Thank you for a PERFECT explanation for my Year 7 class!
@Nordin: It's great to know that this is useful. How old are your year 7's?
Maths with Jay 13 tears old. We do this topic in our enrichment work (early high school in Australia )
@Nordin: Wow! That's impressive. I'm teaching this at a university. (By the way, in the UK, our year 7 students are 11 - 12 years old.)
Nordin Zuber I also need it for tomorrow for an Indonesian Olympiad for grade 9. And I'm also Grade 7.
Thank you, helped me a lot for Security System Theory exam
Brilliant! It's great to know that this is still helping students.
THANK YOU SO MUCH MAM... TOMORROW I'M HAVING MY MATHS EXAM.AND YOU REALLY HELPED ME
Most welcome 😊 Good luck in your exam!
Thank you mam
thankyou , i could not understand this leeson throughout my lectures in uni, then i looked into your video and i'm now so capable of doing this kinda sums.
Excellent! Thank you!
Thank you! Your explanation was very clear and well paced
Thank you so much - that is really useful to know.
Favorite video on Fermat's Little theorem, I like seeing examples.
@Adrian Haro: Many thanks for this feedback
Thank you for this lecture, it's fabulously explained, congrats!
Thanks for this positive feedback.
Ugh math is amazing. Blows my mind every time I learn something new. Awesome video!
Glad you liked it!
Very helpful we love your teaching 😍💝💝
@shiveksha Choubey: Thank you so much...where are you?
@@MathsWithJay I am here 😝
:)
Coud you please explain why remainder does not stay the same for all primes? I tried your example with 2^50 breaking it down with all primes from 17 to 3. For all except 7 and 3 I got remainder 4, which means its composite. But for 7 and 3 I got remainder 1, which means its Prime. I will show it on paper:
2^50 = 2^(10*5) = 1 for P = 11
2^50 = 2^(2*25) = 1 for P = 3
Is there some kinda of rule for choosing primes you are checking for?
I would guess it's because 50 = 5x5x2, 11-1=5x2; 3-1=2? Only just started this topic though so I don't know tbh
This cleared a very annoying gap, thank you!!
Glad it helped!
Hey there, nice video! Do you have a fast way to solve something like '2^550 (mod 551)', or is repeated squaring the only efficient way to solve that?
Because 551 is not prime, you need to use Euler's Theorem here.
Thanks!..It really helped me understand the theorem!!
You are welcome!
5 years later, still useful!
Thank you...I guess mathematics doesn't change much over time
@@MathsWithJay NEVER CHANGES
Thank you so much you explained it better than my teacher
Glad it helped!
Excellent explanation. Quickly understood. 👍👍
Glad it was helpful! 😊😊
This is great explanation. Thank you very much. Keep up the good work
Thank you so much...it's great to see that a video from 2015 is still helping students!
Watching exactly one year after this was published xD. Great video, helped me with my Math Team HW. Thanks a ton! :D
Thanks very much for your feedback
Great video. Cheers from the States! Thank you!
@BaLL1sTiK: Thank you so much!
I love your thick British accent! Much apreesh!
@Nitesh Mishra: lol....thanks
Thanks ❣ solving big examples is tricky but you made it so simple🙏🏻
Most welcome 😊
it took me quite a while to get it,then i came across this video!...the rest is history.:):)
Brilliant! Thanks for letting us know.
Your are great mam..Such a beautiful explanation .. A big Thank You ! Much Love :)
AWESOME!! very helpful. The Aussie accent made me feel much more sophisticated than i actually am hahaha!
Great to hear!....Aussie?
I didn't understand the process of getting frome 16(mod 11) to direct 5(mod 11). Can you help me out here?
Modular arithmetic is like a clock. Once you go around once, it resets back to 1.
Alternatively, we can write 16=(11x1)+5=5mod11.
Thank you so much Ma'am for your nice explanation. It has helped me now. Thank you once again.
@Ibaiahun Muthoh: Thank you so much!
Welcome Ma'am and keep it up Ma'am 👍👍
It was very useful lesson, thanks.
explained it so much better with examples than my book, tysm!
You're very welcome!
Thanks for this great channel! Is there a way of using this theorem to find remainders of ridiculous powers? e.g 3^987654321 mod 11
@ George: It will work if you can write the power as something times 10 plus something, so here 987654321=98765432*10+1
@@MathsWithJay Thank you!
@@MathsWithJay So, as a concept check: 3^(98765432*10+1) = (3^10)^98765432 * 3^1 and under mod 11 that is congruent to 3?
Thank you... the explanation was superb.
Thank you for this great feedback! I'm glad that you found this video useful.
Thanks for your effort, your explination is perfect
Glad it was helpful!
I was having trouble to understand it but you just made it easy.
@Renan: Thank you!
Amazingly clear! Thank you! :)
You're very welcome!
Thanks so much this made the subject more fun and easier
I'm so glad!
Realy helpful for my homework.
Great!
8:27 pls I have a question
Why didn't you divide the 2² by 17
4 < 17
Thank you so much .you taught it so nicely .Thanks Thanks Thanks
Thank you very much!
Great video, helped a lot. Keep up the great work
+Jake Moore: Thanks for your feedback. Much appreciated.
Jay
Maths with Jay
I've been looking around for something particular, maybe you know the answer. How can you do this exact thing but without prime numbers? For example, Find the remainder of 9^342 | 10
+Jake Moore This is a different situation, so think about it in a different way: Start by writing down the first few powers of 9, find the remainders when you divide by 10, and you will see a simple pattern...
supeerb explaination..thnxx a lot
Thank you so much for your positive feedback.
Great explanation, much appreciated.
@Damon: Thank you!
So helpful, thank you very much
You're very welcome!
You are a lifesaver. Thank you!
Glad to hear that!
مادري كيف اوصف لك حبي ياعمري انتي والاكسنت حقتك
You are an awesome teacher : )
Thank you! 😃
Quick question, the formula said a^(p-1) is congruent to 1(mod p) right? So why in the problem of 2^50, did you do it to mod 17?
p = 17 is prime
Grateful, i'm looking your video. Thanks so much
You are welcome!
Good explanation mam....
Thank you so much....Mam.....
Thank you!
Awesome explanation, thank you very much.
Glad it was helpful!
Jay: "4^532... so a really enormous number"
Me *who got 323^5843 on a problem set*: "..pathetic"
The point is that we are considering whether or not the answer can be displayed on a scientific calculator....we start by looking at numbers where the remainder is shown on a calculator, then we look at an example where the numbers can be input without generating an error message, then we look at an example where the calculator cannot evaluate...we are not trying to look at the largest possible example!
@@MathsWithJay I know! I'm just joking using the principal skinner meme format ;)
2:56 hmm that's it :)
WOW , your an amazing teachers, thanks a lot, you made it so easier.Your just the best
Wow, thank you!
Great explanation!
I wonder how you could do this if p is greater than the exponent? Is this theorem still applicable on equations like that?
@NZXTIFY Janne: Thank you! Can you give an example with a larger p value?
Maths with Jay for example 100^11 mod 77837
The mod must be prime
Thank you. Really nice explanation!
Thank you!
thanku maths with jay... love u... plese also make a video on wavy curve(used for solving complex inequalities)
Everything seems so easy after learning it from u xD
@Anonymous AnimeLover: That's really good to know...thank you
Thanks for explanation.😃
:)
Can you add please to your topic regarding shank's square forms factorization.
You madam are a genius!
Thank you!
Thanks for sharing! Was very helpful
Many thanks for your feedback. It is appreciated.
Thank you so much for the clear explanation! Subscribed 😊
Thanks for the sub!
i tried 2^1024 / 131 its perfect thanku..
...Excellent
Very nice explanation !! Can you tell me in detail, what may be the answer for 5^(3125) ?
..that would be an enormous number, with over two thousand digits!
It a fantastic explanation Thank Jay
Glad it helped!
Fantastic 👍👍👌👌
@CHHABI SARKAR: Thank you!
Hi! i have a problem and Im not sure whether its related to this or not. if M^e mod N, N is not a prime number and extremely large, but not larger than e. M is a four digit number, and im trying to so M^e mod N, can I still use this? will the remainder pattern work?
if im doing sth like 1569^2040319087 mod 2203638469. 2203638469 is larger than 2040319087.. should i consider other approach?
Best explanation
Thank you!
Great examples, thank you!
Thank you!
well, you r excellent!! super teaching!!...
Thank you! 😃
What if mod is larger than exponent?
@John: How about trying this for small values and seeing what happens?
Do you divide 50 by 17 ? I think you get 50 = 17 * 2 + 16. Can you clarify please? Thanks
That was so easy to understand. Thank you
Thank you!
Great video! Just wondering, why do you express the theorem as a^(p-1) = 1 mod p instead of as a^p - a = 0 mod p?
It's the same thing!
Thanks alot Jay, you saved me
No problem 👍
Thanks to u i finally understand it THX ❤❤
@Hossam Ali: Thank you!!