It's a Digit World

$2^{2014} \times 5^{2015} \times 10^{2016}$

$= 2^{2014} \times 5^{2014} \times 5 \times 10^{2016}$

$= 10^{2014} \times 5 \times 10^{2016}$

$= 5 \times 10^{4030}$

$= 5 \times 1\underbrace{000...0000}_{\text{4030 zeroes}}$

$= 5\underbrace{000...0000}_{\text{4030 zeroes}}$

So the total number of digits is the $4030$ zeroes plus the digit five in the front which gives us $4030 + 1 = \boxed{4031}$

nice problem

My solution is different and it is easy too..
Powers are in arthimetic progression then take any one or two examples to understand it.
2^3 ×5^4 ×10^5 = 5×10^8 =9
2^4 ×5^5 ×10^6 = 5×10^10 =11
By both examples we got it no.of digits are the addition of power of 5 and 10. Same we ll do in that case and get our answer.
2015+2016= 4031 Ans: