How many positive integers are there such that ⌊√x⌋=10?

My answer was 11 but somehow that was wrong.

if x <= 100 < 121 then ⌊√x⌋=10

I don't get what you mean Steve so that means that there are 50 positive integers where it equals 10?

To determine the number of positive integers satisfying the equation ⌊√x⌋ = 10, we need to understand what the floor function, ⌊x⌋, does.

The floor function gives the largest integer that is less than or equal to a given real number x.

In this case, we are looking for positive integers x such that ⌊√x⌋ = 10. It means that the floor of the square root of x should be equal to 10.

To find these numbers, we can examine the values of √x that would yield a floor value of 10. Since the floor function rounds down, we are interested in the range of values of √x that is greater than or equal to 10 but less than 11.

So, we need to solve the inequality: 10 <= √x < 11.

To do this, we can square both sides of the inequality: (10)^2 <= (√x)^2 < (11)^2.

This simplifies to: 100 <= x < 121.

Since x has to be a positive integer, we need to count the positive integers from 100 to 120 inclusive.

The number of positive integers in this range can be calculated as:

Number of positive integers = (120 - 100) + 1 = 21.

Therefore, there are 21 positive integers x that satisfy the equation ⌊√x⌋ = 10, not 11.

Hence, your answer of 11 was incorrect.