+1439 Let f(x) = \sqrt{x}
Find the largest three-digit value of x such that f(x) is an integer.
+1926 Let's note something for this problem first.
We are basically taking a look at the squares of numbers, and eeing the largest 3 digit number.
Thus, let's test some integers. We find that
\(31^2 = 961 \\ 32^2 = 1024 \)
This shows that 31 is the largest value squared.
Thus, 961 is our answer.
So we have \(x=961\)
Thanks! :)
