+118 Let
\(f(x) = \sqrt{x - \sqrt{x - \sqrt{x - \sqrt{x - \dotsb}}}}\)
Find the largest three-digit value of such that is an integer.
Letting y=f(x), we get that y(y+1)=x, and y(y+1) is a three-digit number.
Solve for y
+195 Analyzing the function:
This function defines f(x) as the square root of x minus a nested sequence of square roots that keep getting smaller. Intuitively, as x decreases, the nested square roots will also decrease, bringing f(x) closer to an integer.
Finding a bound:
For f(x) to be an integer, x - sqrt(x - sqrt(x - ...)) must be a perfect square. Let the innermost square root be y. We can rewrite the function as:
f(x) = sqrt(x - y)
Squaring both sides:
x - y = f(x)^2
Since we want the largest possible three-digit x, let's consider the smallest possible value f(x) can take. The smallest perfect square greater than 10 (a three-digit number) is 121. So, let's assume f(x) = 11.
Substituting:
x - y = 11^2 = 121
This tells us that x needs to be at least 121 more than the innermost square root (y) to be a perfect square.
Finding the largest three-digit x:
We know x must be greater than 121 + y. Since y is a square root and squares are always non-negative, the largest possible value of y for a three-digit x would be the square root of the largest three-digit perfect square, which is 9^2 = 81.
Therefore, x must be greater than 121 + 81 = 202.
Checking values:
We can now check the largest three-digit perfect squares less than 202:
196 (14^2) - When plugged into the function, it results in a non-integer value.
169 (13^2) - This value works! f(169) = 13, which is an integer.
Therefore, the largest three-digit value of x such that f(x) is an integer is x = 169.
