What will be remainder when 19 ¹⁰⁰ is divided by 20?

(9¹⁹ + 6 ) / 8 = [( 8 + 1 )¹⁹ + 6] / 8
? Required remainder = ( 1¹⁹ + 6 ) / 8

? Remainder = 7

We know that when m is a odd number ( x^m + a^m ) is divisible by ( x + a ).
? Each one is divisible by (41 + 43).
So Common factor = (41 + 43).

Remainder of 4¹⁰⁰⁰/7 = (4²)⁵⁰⁰/7 = (16)⁵⁰⁰/7
= 2⁵⁰⁰/7
= Remainder of 4 x 8¹⁶⁶/7
= 4

We know that, ( x^m - a^m ) is divisible by ( x + a ), for even value of m.
? 17²⁰⁰ - 1²⁰⁰ is divisible by 17 +1
? 17²⁰⁰ - 1 is divisible by 18.

Hence when 17²⁰⁰ divided by 18 the remainder is 1.

We know that, a number is divisible by 11 when the sum of the difference between the sum of its digit at even places and sum of its digit at odd places is either 0 or the difference is divisible by 11.

For 93455120
Sum of even places = 3 + 5 + 1 + 0 = 9
Sum of odd places = 9 + 4 + 5 + 2 = 20

There difference = 20 - 9 = 11
which is divisible by 11 so 93455120 is divisible by 11.

Let 2³² = x and let ( 2³² + 1 ) = ( x + 1 )
( 2⁹⁶ + 1 ) = ( x³ + 1 ) = ( x + 1 )( x² -x + 1 )
Which is clearly divisible by n as ( x + 1 ) is divisible by n.

xⁿ - yⁿ is divisible by x + y and x - y for all n.

7¹² - 4¹² is divisible by 7 + 4 and 7 - 4
? 7¹² - 4¹² is divisible by 11 x 3 = 33

( 2³ - 2) = 6 is the largest natural number that divides ( a³ - a) for every number a.

We can check divisibility of 19⁵ + 21⁵ by 10 by adding the unit digit of 9⁵ + 1⁵ which is equal to 9 + 1 = 10.
So it must be divisible by 10.

Now, for divisibility by 20 we add 19 and 21 which is equal to 40. So, it is clear that it is also divisible by 20.

So 19⁵ + 21⁵ is divisible by both 10 and 20

6x² + 6x = 6x( x + 1 ) which is clearly divisible by 6 and 12 as x( x + 1 ) is even.