The square root of $x$ is greater than 3 and less than 4. How many integer values of $x$ satisfy this condition?
Because $4 > \sqrt{x} > 3$, we know that $16 > x > 9$.  Thus, the integers from 10 to 15 inclusive satisfy this inequality, which means $\boxed{6}$ integers satisfy the condition in the problem.
6
Level 4
Prealgebra
Problem: The square root of $x$ is greater than 3 and less than 4. How many integer values of $x$ satisfy this condition?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the integer values of $x$ that satisfy the condition that the square root of $x$ is greater than 3 and less than 4, we start by translating the inequality involving the square root into a more manageable form. 

First, we express the condition mathematically:
3 < √x < 4.

Next, we square all parts of the inequality to eliminate the square root. This gives us:
3^2 < x < 4^2,
which simplifies to:
9 < x < 16.

Now, we need to identify the integer values of $x$ that fall within this range. The integers greater than 9 and less than 16 are:
10, 11, 12, 13, 14, and 15.

Counting these integers, we find there are 6 values: 10, 11, 12, 13, 14, and 15. 

Thus, the total number of integer values of $x$ that satisfy the original condition is 6.<|END_OF_LATENT|> The final answer is $6$.