19 ⁵ + 21 ⁵ is divisible by

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

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

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

Let the number be x and (15-x)
Then, x² + (15 - x)²= 113
? x² - 15x + 56 = 0
? (x-7) (x-8) = 0
? x = 8 or x = 7
So, the numbers are 7 and 8