Find a positive integer that is divisible by 14 and has a square root between 25 and 25.3.
Let's call the positive integer we're looking for x. We know that x is divisible by 14, so it must be divisible by 2 and 7. Right, so it must be an even number. Also, we know that $\sqrt{x}$ is between 25 and 25.3. So, $25^2<x<25.3^2$. We also know that x is an even number. So, it must be divisible by 2. So, x could be 626, 628, 630, 632, 634, 636, 638, or 640. We also know that x is divisible by 7. So, $x=630$.

