Wow, i spend 2 hours trying to understand my lecturer's notes on modular exponentiation, even went through so many web tutorials, and still wasn't clear. This one single video just made me understand the whole concept in 10 min. Great!!! Thank you so much for the clear and concise explanation!!!
Hello Neso Academy,I really highly applaud you👏 for making this modular exponentiation a simply but perfectly explained one😍.❤From Bangladesh.And also my endless thanks🥰☺
For those who are still confused - 23³ mod 30 = (23 x 23 x 23) mod 30 In the prev video, we saw the property - (a mod n x b mod n) mod n = (a x b) mod n Using this property of modular arithmetic from the previous video, we get = [23 mod 30 x 23 mod 30 x 23 mod 30] mod 30 Now, since we know 23 mod 30 = 23, thus (23 x 23 x 23) mod 30 = 12167 mod 30 = 17. Alternatively, we can also substitute 23 mod 30 with -7 mod 30 which makes the calculation simpler and which is also done in the video. = [-7 mod 30 x -7 mod 30 x -7 mod 30] mod 30 Again using the same property of modular arithmetic, = (-7 x -7 x -7) mod 30 = (-7)³ mod 30 = (49 x -7) mod 30 Again using the property - = [49 mod 30 x -7 mod 30] mod 30 = [19 mod 30 x -7 mod 30] mod 30 = (19 x -7) mod 30 = -113 mod 30 We see 49 x -7 mod 30 would give -343 mod 30 = 17 (since -343+30x12 = 17, to make -343 positive we had to add 30 twelve times which yielded 17). But 49 was simplified to 19. The properties of modular arithmetic have been used implicitly to make the calculation easier.
the thing is not with your professor, it's that when you was in the school you was tired or not consentrated so that it came difucile by the way i am not a teacher
@@I_just_love_psyou don't have to have good teaching skills to be a professor. Most professors are really bad at teaching because they aren't trained for that. Nesoacademy has valuable "teachers"
In the calculation there are errors 23^3 mod 30, in the fifth step its -343 mod 30 and error in the calculation 11^7 mod 13 there is an error in the fourth step
@@carterschmidt7411 if adding x in mod x lets the number be the same, then subtracting x should also keep the number the same 49 - 30 = 19, and 19 - 30 is -11 so 49 = -11 mod 30
Here's an informal way to do 49 * -7mod30 = -113mod30 49 * -7 = -343 Thus: -343mod30 and now we have to simplify/reduce If you recall from neso's earlier video on modular arithmetic in the part covering congruence: If a is congruent to b(modm) then a = km+b In practice this means that for any integer 'k': -343mod30 = (k(30) + (-343))mod30 so choosing k = 7: -343mod30 = (210 + -343)mod30 = -133mod30 However, I think it's easiest to just choose largest k such that km is still less than |b| i.e. choose largest k so k(30) < 343 k=11 -343mod30 = (330 + -343)mod30 = -13mod30 Hope that helps.
can I do like this : 31^500 mod30 =(30+1)^500 mod 30 ; [30^500+2*30*1+1^500] =mod30 ; =30^500 mod 30+60 mod30+ 1mod30 0+0+1mod 30 =1mod30; 1/30 =Q=0+R=1 =1
@@MrBlancify That is a very relevant solution. Thanks Can you tell how to know that there are only 02 possible solutions of the 23 mod 30? Can’t there be more solutions?
@@MaheshKumar-vi7pi Techincally infinite solutions. It could be -277 for example, 23 = 30*10-277. But the solutions we want in modulo are ones closest to zero.
so, he did 49 mod 30 first, which equals 19, and then he multiplied 19and -7 and got -133 honestly, it took me a bit to figure out, but since he doesnt want us to use a calculator, he just did it this way
1) There are a lot of good students who support these organization by donating some money. 2) This academy also make some money from UA-cam. 3) There are some paid courses that are only available on nesoacademy website from there they also make some money. That's how they get money to eat.
49*(-7) = -343
49 mod 30 is 19 ........and then you multiply 19 by -7 and you get -133.
Wow, i spend 2 hours trying to understand my lecturer's notes on modular exponentiation, even went through so many web tutorials, and still wasn't clear. This one single video just made me understand the whole concept in 10 min. Great!!! Thank you so much for the clear and concise explanation!!!
hello how did 49*-7mod30 become -133mod30
He use reminder value of (49 ÷ 30 = 19) and 19 * 7 = 133
Hello Neso Academy,I really highly applaud you👏 for making this modular exponentiation a simply but perfectly explained one😍.❤From Bangladesh.And also my endless thanks🥰☺
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For those who are still confused -
23³ mod 30
= (23 x 23 x 23) mod 30
In the prev video, we saw the property -
(a mod n x b mod n) mod n =
(a x b) mod n
Using this property of modular arithmetic from the previous video, we get
= [23 mod 30 x 23 mod 30 x 23 mod 30] mod 30
Now, since we know 23 mod 30 = 23, thus (23 x 23 x 23) mod 30 = 12167 mod 30 = 17. Alternatively, we can also substitute 23 mod 30 with -7 mod 30 which makes the calculation simpler and which is also done in the video.
= [-7 mod 30 x -7 mod 30 x -7 mod 30] mod 30
Again using the same property of modular arithmetic,
= (-7 x -7 x -7) mod 30 = (-7)³ mod 30
= (49 x -7) mod 30
Again using the property -
= [49 mod 30 x -7 mod 30] mod 30
= [19 mod 30 x -7 mod 30] mod 30
= (19 x -7) mod 30
= -113 mod 30
We see 49 x -7 mod 30 would give -343 mod 30 = 17 (since -343+30x12 = 17, to make -343 positive we had to add 30 twelve times which yielded 17). But 49 was simplified to 19. The properties of modular arithmetic have been used implicitly to make the calculation easier.
Thanks, This helped than the video
So you used commutative property and taken both A= 23, B=23.
How?
thanksssssssss
thanks broooo
Great comment :)
Thank you very much. Your explain is much easier understand than my text book.
Thank you, you explained it much simpler than my learning material!
This video gave me a booster,.. thank you 😎🙏💐
I'd appreciate if u take the next dose too
Thank you, I didn't catch it well the modular exponentiation, but with ur video I uderstand it!!!
You have explained it in the most easiest version. Thank you.
hello how did 49*-7mod30 become -133mod30
Thanks sir 🙏 the way you explain the things is splendid 👌 👏
I love the way you explain!!! Way better than my current professor!!!😁💕
the thing is not with your professor, it's that when you was in the school you was tired or not consentrated so that it came difucile
by the way i am not a teacher
@@I_just_love_psyou don't have to have good teaching skills to be a professor. Most professors are really bad at teaching because they aren't trained for that. Nesoacademy has valuable "teachers"
In the calculation there are errors 23^3 mod 30, in the fifth step its -343 mod 30 and error in the calculation 11^7 mod 13 there is an error in the fourth step
it not a mistake he just simplified the expression, instead of 49*-7 mod 30 he did (49mod30 / which equal to 19)*-7mod30, result is exactly the same
Can you provide us the whole syllabus of this course or the length of this course with roadmap
Good video, but you need to put parentheses around negative bases to avoid ambiguity. Otherwise, it is parsed -(7^3).
I was just about to say that :)
how can 49*7= 133 instead of 343???
exactly!!!!!!!!
@@sammuriithi1313 he skipped the part that was supposed to be (49 mod 30)=19 multiplied by ( -7 mod 30). so then (19 x -7 ) mod 30 = -133
This is really helpful.
He use reminder value of (49 ÷ 30 = 19) and 19 * 7 = 133
saving me thank so very much
For example if the mod is 284 so how can you calculate it ? Every time you made it as 1 of course easy to calculate
Great 😍
Thank you so much buddy ❤
at 4:11 i think 49*-7 = -343
Yes you are right 👍
@@Utkarshkushwaha-ld8xh All the best for your exam.
49 * -7 (mod 30) = -11 * -7 (mod 30) = 77 mod 30 = 17
How did you get -11?
@@carterschmidt7411 if adding x in mod x lets the number be the same, then subtracting x should also keep the number the same
49 - 30 = 19, and 19 - 30 is -11 so 49 = -11 mod 30
atleast yours is better than that -133 i didnt get that calculations
@@benjaminrutto8365 yeah, I too didn't understand that
49×7 = 343 . I don't know which math you are teaching but explain it...
even i got confused for a min.
same
30×11= 330
So reminder is -13
17 Answer is correct
49-30= 19 (49 mod 30)
19*7 = 113
@@princepatel6645 crt bro
Hello sir can you do video on RSA algorithm plz 🙏🏻
Thank you so much
Sorry guys? -133 mod 30 = -13. COuld you please explain? Thanks
30*4=120. -13 is left
umm just a update 49 * -7=-343
49 mod 30 is 19 ........and then you multiply 19 by -7 and you get -133
He use reminder value of (49 ÷ 30 = 19) and 19 * 7 = 133
Thank you so much!!!!
Thank you very much 😊
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How is 23 mod 30 is equal to -7 mod 30 ?
you can subtract 23 by 30 which is -7 and -7 mod 30 gives the same result
And for bigger like 1816^13 mod 2437
4:10 -7 * 49 = 133 ?? how
But what if it is 2015^17mod 3233 ? My problem is still didn't solve
it only works if the number gets small if u cant get the number small both number and power are big u cant solve it this way
example 644^21 mod 697
Writing -1^329 triggers my OCD, please write (-1)^329 thanks ;)
-7 *-7 =+49?
49 * -7 mod 30 = -113 mod 30 how it came can any1 explain?
Here's an informal way to do 49 * -7mod30 = -113mod30
49 * -7 = -343
Thus: -343mod30 and now we have to simplify/reduce
If you recall from neso's earlier video on modular arithmetic in the part covering congruence:
If a is congruent to b(modm) then a = km+b
In practice this means that for any integer 'k':
-343mod30 = (k(30) + (-343))mod30
so choosing k = 7:
-343mod30 = (210 + -343)mod30 = -133mod30
However, I think it's easiest to just choose largest k such that km is still less than |b|
i.e. choose largest k so k(30) < 343
k=11
-343mod30 = (330 + -343)mod30 = -13mod30
Hope that helps.
@@maxwellconniff1189 Hi, where did you get 7 = k? From -7? How did it change to +7 then?
Thanks.
This was mistake just leave it and do calculation on your own 🙃
Thumbs down. No proper explanation on how 47 * -7 = 133. Garbage
-7 * 49 mod 30 = -133 mod 30
49 mod 30 = 19
So 19 * -7 mod 30 = -133 mod 30
how does -2 seven times is equal to -128?
-2 * -2 * -2 * -2 * -2 * -2 *-2 = -128
maybe you were thinking he -2*7 ?
49*-7 = 133 ? am i tripping or sth?
49mod30=19, 19*-7 = 133
He needs a calculator to do 49*7😂
how did you get -7????? for 23
-7 === -7 +30 === 23 mod 30
49 x 7 = 343 ?
2 pow 343 mod 142 will be
Not specified perfectly. Very bad calculation. Calculations not explained
You Should Explain The Calculations In a Better Way, Rest Of It is Good!
can I do like this : 31^500 mod30 =(30+1)^500 mod 30 ; [30^500+2*30*1+1^500] =mod30 ; =30^500 mod 30+60 mod30+ 1mod30
0+0+1mod 30 =1mod30; 1/30 =Q=0+R=1 =1
38*8 mod 63 how can find
How to solve (5)^15 mod 23 ?
Please Upload Steam and Block Ciphere
I think it is available on website.
Good luck
sir do for 176 power 23 mod 187
Why?
@@adinapunyobanerjee9481 Because it seems this is not foolproof
Can anyone solve 2^(62)mod77=?
How is -128 mod 13 = -11 mod 13
how can 23 mod 30 be -7 ??
Sir how got -13
Sir please next video upload kijiye
can anyone explain to me logic behind having -7 as a result of 23 mod 30
23 = 30*0 + 23 or you can have 23=30*1-7.
@@MrBlancify That is a very relevant solution. Thanks
Can you tell how to know that there are only 02 possible solutions of the 23 mod 30? Can’t there be more solutions?
@@MaheshKumar-vi7pi Techincally infinite solutions. It could be -277 for example, 23 = 30*10-277. But the solutions we want in modulo are ones closest to zero.
@@MrBlancify Thanks Man. Great explanation.
@@MrBlancify -7 is closest to 0 than 23 itself, hence we go with -7? Right?
11K views means 11K person got benefited from your lecture so almighty will arrange for your eating
It's 16K views now.
@@陈妖精 its now 185,078
191 k now
find the value of 6^24 mod 35 using exponentiation
How sir
@@matheshwaran4028I think that if you write 6^24 = 6^(2*12)= (6^2)^12=36^12
Since 36 = 1(mod35)
Then 36^12(mod35) = 1^12(mod35) =1(mod35)
example 1-------->>>> (49*(-7))=133 how?
so, he did 49 mod 30 first, which equals 19, and then he multiplied 19and -7 and got -133
honestly, it took me a bit to figure out, but since he doesnt want us to use a calculator, he just did it this way
What answer if 14^27 mod 55 !!?????? You solve easy example try hard And prove your mind !...
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kuch samjh naa aaya
Sir, if you are teaching for free. Then how did you get money to eat.
1) There are a lot of good students who support these organization by donating some money.
2) This academy also make some money from UA-cam.
3) There are some paid courses that are only available on nesoacademy website from there they also make some money.
That's how they get money to eat.
They earn money more than your expectations 🙃
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We have calculator
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@ 😂good luck