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?

Show Hint

y is only an integer when x is even, or when x is an odd perfect square (since then \(\sqrt{x}\) is a whole number); tabulate y for x = 1,2,4,6,8,9,... and pick out the values falling in the three-digit and four-digit ranges.
Updated On: Jul 20, 2026
  • 1220
  • 2450
  • 3240
  • 3880
  • 4280
Show Solution

The Correct Option is D

Solution and Explanation

Approach this by bracketing x directly using logarithms/digit-count instead of listing every value.
We need \(y=x^{x/2}\) to have exactly 4 digits, i.e. \(1000\le y\le9999\), and separately exactly 3 digits, i.e. \(100\le y\le999\).
Since \(y\) grows extremely fast with \(x\) (it is a "power tower"-like function), only consecutive small integers need checking, and \(y\) must be rational, which restricts \(x\) to even numbers or odd perfect squares as argued.
Testing consecutive valid \(x\): \(x=4\to16\), \(x=6\to216\), \(x=8\to4096\), \(x=9\to19683\).
Between \(x=6\) (216, 3-digit) and \(x=8\) (4096, 4-digit) the jump skips straight over any other 3-digit or 4-digit candidate (there is no valid x giving a value strictly between 216 and 4096, since x=7 is invalid and there's no integer x between 6 and 8). So 216 is the unique 3-digit value and 4096 is the unique 4-digit value.
Difference:
$$4096-216=3880$$\[\boxed{3880}\]
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