Solution:
$$\eqalign{ {7^{6n}} - {6^{6n}} \cr = {7^6} - {6^6} \cr = {\left( {{7^3}} \right)^2} - {\left( {{6^3}} \right)^2} \cr = \left( {{7^3} - {6^3}} \right)\left( {{7^3} + {6^3}} \right) \cr = \left( {343 - 216} \right) \times \left( {343 + 216} \right) \cr = 127 \times 559 \cr = 127 \times 13 \times 43 \cr} $$
Clearly, it is divisible by 127, 13 as well as 559

Solution:
aⁿ + bⁿ is always divisible by (a + b) when n is odd.
So,
(2222⁵⁵⁵⁵ + 5555²²²²) is always divisible by (2222 + 5555) = 7777
And 7777 is multiple of 7,
so (2222⁵⁵⁵⁵ + 5555²²²²) is divisible by 7

Solution:
= (a² + b² + c² - ab - bc - ca)
= (a + b + c)² - 3 (ab + bc +ca)
= (7 + 5 + 3)² - 3 (35 + 15 + 21)
= (15)² - 3 × 71
= (225 - 213)
= 12

Solution:
Every real number is a rational number

Solution:
Each prime number have to distinct factor number itself and 1.

Solution:
Clearly, 19 is a prime number.

Solution:
When two two digit numbers are added and the resultant value is a three digit number. It means there must be a carry over (i.e. The sum of unit digits be greater than 9. Similarly, the sum of the tens digit is also greater than 9) Required numbers = 64 + 46 = 110 Option D is correct