Question:hard

If 'y' is a number such that \( y = x^{x/2} \), where x is a positive integer, what is the difference between the largest possible four-digit value of y and the smallest possible three-digit value of y?

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Only check values of x where y comes out as a whole number: even x always works, odd x works only when x is itself a perfect square.
Updated On: Jul 21, 2026
  • 1220
  • 2450
  • 3240
  • 3880
Show Solution

The Correct Option is D

Solution and Explanation

The rule \(y = x^{x/2}\) only gives a clean whole number for y in two situations: when x is even, so the power x/2 is a whole number, or when x is odd but is itself a perfect square, so the leftover square root cancels out. Let's work through both families separately, since that's a more direct way to see why only one number fits each digit range.

  1. Even x: x = 2, 4, 6, 8, 10, ... give y = 2, 16, 216, 4096, 100000, ... These values jump very fast, so they skip over most digit ranges entirely. Checking each one: 2 and 16 are too small, 216 lands in the three-digit range, 4096 lands in the four-digit range, and 100000 already has six digits.
  2. Odd x that are perfect squares: x = 1, 9, 25, ... give y = 1, 19683, and a huge number for x = 25. Only x = 1 gives a small value, and it's just a one-digit number; x = 9 already jumps to a five-digit value, so no odd x contributes anything in the three-digit or four-digit range.

So across both families, the only three-digit value of y is 216, from x = 6, and the only four-digit value of y is 4096, from x = 8. Since each range has exactly one qualifying value, that value is automatically both the largest and the smallest in its own range.

Let's summarize:

  • Even x always gives a whole-number y, and odd x gives a whole-number y only when x is itself a perfect square.
  • Checking both families shows 216 is the only three-digit y and 4096 is the only four-digit y.

The required difference is 4096 - 216 = 3880.

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