How to Find the Remainder | Number Theory

How to Find the Remainder | Number Theory

Join this channel to get access to perks:

Topics covered in this video
what is remainder theory
the concept of negative remainder
remainder theorem in number system
remainder theorem
remainder
number theory
michael penn
mathematics
math
how to find the remainder - number theory
how to find remainders easily
how to find remainders
find the remainder with the help of congruence
find the remainder
fermat’s little theorem
division
abstract algebra

[devondevon]

Answer 1
A different approach
9^1990
can be written as 3^2^1990 =
3^ 3980, which can be written as
3^5)^756
243^ 756
243 is congruent to 1 mod (11) since 11 divides 242 or 242 is congruent to 0 mod 11
hence 243 can be replaced by 1
1^756, which is the same as 1. hence the remainder is 1

[robertradutu]

Hello, what is the name of the app that you are using in these videos to draw?

[mathrapper]

In my opinion, it is easier to do something like this: 9 power 1 (any number) is equal to a remainder, repeat until you get the remainder that matches that of 9 power 1
Example
9 power 1 equals 9 (mod 11)
9 power 2 equals 4 (mod 11)
9 power 3 equals 3 (mod 11)
9 power 4 equals 5 (mod 11)
9 power 5 equals 1 (mod 11)
9 power 6 equals 9 (mod 11)
9 power 1, 6, 11 and so on have the same remainder.

[sy]

Thank you for explaining by using unusual method. For me, I think using [9^4≡5 (mod 11) and] 9^5≡1 (mod 11) is easier to solve this problem.
・・・・・・　From 5|1990, 9^1990 is also ≡1 (mod 11)

[jarikosonen]

9^5mod11=1 is the 1st giving remainder one and 9^1990=9^(5*398), so that would give remainder =1 as well. Looks wrong thing to do. Because the cycle is 5 when the remainder repeats, like 9^10mod11=1, 9^15mod11=1, etc

[brinzanalexandru]

Also you could've use Fermat little theorem

[mhmmdmmmdov]

Hello. I think this question is very easy. Because if you use fermat's little theory, 9^10=1 (mod 11)
Then, 1990=0 (mod 10) and you can say that 9^1990=1 (mod 11)

[anandharamang]

9*9= 81, 81 (mod 11) = 4, then 9^3 (mod 11) = 4*9 (mod 11)= 3, then 3*9 (mod 11)= 5, then 5*9 (mod11) = 1, so 9^5 (mod 11)= 1, 1990 is divisible by 5, so, remainder must be 1. It is simple and easy even for any powers to calculate in this way.

[Chettick]

GOOD DAY. I AM THE DESIGNER OF MORFOLOGIC TEOREM MCHETTICK. A MATHEMATICAL PROBLEM THAT YOU MUST BE SEE AND SOLVE... CAN YOU SOLVE?

[stkhan]

Since 11 is prime
phi(11)=10 ; phi-- euler number
(9)^10k=1(mod11) ; 1990=10k

[harshitverma]

Please give a simple method for a high school student to understand this type of questions

[CenturionDobrius]

- little bit complicated, as there are at least two extremely simple solutions :
1. Fermat's little theorem,
2. and a fact that : 9 power 5 mod 11 = 1, (9 power 5) power 398 = 1 mod 11.
- a friendly request : try to be less verbose in explanation and writing :)

[stkhan]

Since 11 is prime
phi(11)=10 ; phi-- euler number
(9)^10k=1(mod11) ; 1990=10k