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🥰☺
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!!!
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.
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
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"
@@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
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
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.
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
@@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.
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