If $x$ is a positive multiple of 8 and $x^2>100$, but $x<20$, what is $x$?
This question gives multiple clues on what $x$ could be. Because $x^2>100$, we know that $x$ must be greater than 10, but $x$ is also less than 20. So $x$ is a multiple of 8 that is between 10 and 20. The only value that fits this description is $\boxed{16}$.
16
Level 1
Prealgebra
Problem: If $x$ is a positive multiple of 8 and $x^2>100$, but $x<20$, what is $x$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for \( x \), we first identify the positive multiples of 8 that are less than 20. The multiples of 8 are 8, 16, 24, etc. Since \( x < 20 \), the relevant multiples are 8 and 16.

Next, we need to check which of these values satisfies the condition \( x^2 > 100 \). 

Calculating \( 8^2 \):
\( 8^2 = 64 \), which is not greater than 100.

Calculating \( 16^2 \):
\( 16^2 = 256 \), which is greater than 100.

Since 16 is the only value that meets both conditions of being a positive multiple of 8 and satisfying \( x^2 > 100 \), we conclude that the solution to the problem is \( x = 16 \).<|END_OF_LATENT|> The final answer is $16$.