If 'a' is a natural number, then the largest number dividing ( a ³ - a) is

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

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.

19¹⁰⁰ = ( 20 -1 )¹⁰⁰
? ( -1 )¹⁰⁰ / 20
? 1 / 20
? Required Remainder = 1

(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).

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.

When N is a natural number, then there is only one possible case that N, N + 2, N + 4 are prime numbers.

When N = 3, then N, N + 2, N + 4 = 3, 5, 7 all are primes.

Taking p = 11
2^p - 1 = 2¹¹ - 1
= 2047

Since 2047 is divisible by 23 it is not prime. Thus, required least positive prime number is 11.

Let the two part be x and (24 - x),
Then, 7x + 5(24 - x) = 146
? 7x + 120 - 5x = 146
? 2x = 26
? x = 13
? First part = x = 13