Math Interview Tricks: From Stanford, Harvard, Cambridge to Oxford University

[@billv39]

I find it interesting that you can repeatedly nest the left side and get the same x=6 result. For example √(30 + √(30 + √(30+x))) = x. There might be some bigger takeaway that I'm missing. Beyond that, it's now apparent how to create similar problems, ex. √(20 + √(20+x)) = x gives x=5. Or √(132 + √(132+x)) = x gives x=12.

[@superacademy247]

That's a great observation! You're right, there's a pattern there. 🤔It's amazing how you found that pattern. You can create a whole family of similar problems! 🤯

[@billv39]

@@superacademy247 Once you realize what's going on, you can easily create all sorts of crazy problems. For example √(31 + √(21 + √(10+x))) = x has a solution of x=6. Also I see that it's possible to unwind ...√(30 + √(30 + √(30+x))) = x where there are an infinite number of nested √(30+'s. Just substitute the right side, x, for everything on the left except the outermost √(30 + ). That gives you √(30 + x) = x, which becomes x^2 - x -30 = 0. And x=6 like before.

[@prollysine]

x^4-60x^2-x+870=0 , (x-6)(x+5)(x^2+x-29)=0 , only solution x=6 ,

[@wasimahmad-t6c]

X= Squairooth 30

[@yangchong4680]

Stupid method