Question

Let the number of digits of `2^2005` be x and that of `5^2005` be y then the possible value of x+y=?

Answer 1

`2005`

Explanation:

Note that "number of digits" of `n` is essentially `floor(log_10 n) + 1`

So:

`x = floor(log_10 2^2005) + 1 = floor(2005 log_10 2) + 1`

`y = floor(log_10 5^2005) + 1`

`color(white)(y) = floor(2005 log_10 5) + 1`

`color(white)(y) = floor(2005(1 - log_10 2)) + 1`

`color(white)(y) = floor(-2005(log_10 2)) + 2006`

`color(white)(y) = -ceil(2005 log_10 2) + 2006`

So:

`x + y = floor(2005 log_10 2) - ceil(2005 log_10 2) + 2006 = 2005`