Maths hard problem

$\begin{aligned} 2^{2999} \times 5^{3002} & = 5^3 (2 \times 5)^{2999} \\ & = 125 \times 10^{2999} \\ & = 125\underbrace{00000 \cdots 00000}_{\text{Number of 0s}=2999} \end{aligned}$

Therefore the sum of digits is $1+2+5 = \boxed 8$ .

2²⁹⁹⁹ x 5³⁰⁰²

= 2²⁹⁹⁹ x 5²⁹⁹⁹ x 5³

= (2 x 5)²⁹⁹⁹ x 5³

= (10)²⁹⁹⁹ x 125

= 12500...00 ( The '...' represents the rest zeroes)

Sum of digits = 1+2+5+0+..+0

= 8

2^2999 ×5^3002 can be written as 2^2999× 5^2999×5^3 Which is equal to 10^2999 ×5^3 Evaluating it: 125 ×10^2999 So the sum of digit is 1+2+3+5 is 8