Floor it!

The question asks for the number of values for which $\sqrt{x}$ is greater or equal to 10 and less than 11. In order for this to occur, $10\le\sqrt{x}<11$ . So, we have $100\le{x}<121$ , which gives us $\boxed{21}$ positive integers.

De acordo com a definição temos:

√100 ≤ √x < √121 √100 ≤ √x ≤ √120

Daí,

120 - 100 + 1 = 21

$⌊\sqrt {x}⌋=10$ means the the roots of the values of x have to be $\geq 10$ but $<$ 11. So we get the following inequalities:

$\sqrt {x} < 11 \quad and \quad \sqrt {x} \geq 10$

Solving these we get:

$x < 121 \quad and \quad x \geq 100$

So the range for x that satisfies those conditions is [100, 121). Since 100 is inclusive and 121 is not, we can simply do 121 -100 = 21 to get our answer.

⌊ x√ ⌋ = 102 ≤ x2 < 112 ⌊ x√ ⌋ = 100 ≤ x2 < 121

thus, [100, 121)

and there are 21 integers in this range

[Staff Edit- This user has been removed from Brilliant for plagiarism]

⌊ $\sqrt{x}$ ⌋ = $10^{2}$ ≤ $x^{2}$ < $11^{2}$

⌊ $\sqrt{x}$ ⌋ = 100 ≤ $x^{2}$ < 121

thus, [100, 121)

and there are 21 integers in this range