$PI$ is a two digit number and $PIE$ is a three digit number, where each letter is a different digit from 0 to 9 ($P\ne0$). If $\sqrt{PI}+E=\sqrt{PIE}$, what is the value of $P+I+E$?

I feel you. I started by squaring both sides to get rid of the radical. Then it became a matter of testing digits that satisfy the equation. Not too bad if you organize.

Yeah, the notation can be confusing. I wish they'd write it as a number with a bar over it or something.

Plugging in is actually a valid strategy here, but you can narrow it down by noting that E must be small because the difference between sqrt(PIE) and sqrt(PI) is E. So E is likely 1, 2, or 3.

Maybe try expressing PIE as PI*10 + E? But PI is two-digit, so PIE is not simply 10*PI + E because the digits shift. Actually PIE = 100P + 10I + E, while PI = 10P + I. So PIE = 10*(10P+I) + E = 10*PI + E. Yes! That's the key. Then sqrt(10*PI + E) = sqrt(PI) + E. Let x = sqrt(PI). Then x^2 = PI, and sqrt(10x^2 + E) = x + E. Square: 10x^2 + E = x^2 + 2Ex + E^2 => 9x^2 - 2Ex + (E - E^2) = 0. That's a quadratic in x. Might be easier to test integer x.