Question

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?

• A. 1220
• B. 2450
• C. 3240
• D. 3880

Hint:

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.

Answer 1

The Correct Option is D

Step 1: Understand what makes y a whole number.
We have \( y = x^{x/2} \), where x is a positive integer. For y to be a definite four-digit or three-digit value, y itself should be a whole number, not some irrational decimal.
When x is even, x/2 is a whole number, so y is always a whole number. When x is odd, x/2 is a fraction, and y is a whole number only when x itself is a perfect square, so the leftover square root cancels out neatly.

Step 2: List out the whole-number values of y for small x.
x = 1: \( y = 1^{0.5} = 1 \)
x = 2: \( y = 2^{1} = 2 \)
x = 4: \( y = 4^{2} = 16 \)
x = 6: \( y = 6^{3} = 216 \)
x = 8: \( y = 8^{4} = 4096 \)
x = 9: \( y = 9^{4.5} = 9^4 \times 3 = 19683 \)
x = 10: \( y = 10^{5} = 100000 \)
Between these, x = 3, 5, 7 give irrational values and are skipped, since none of them is a perfect square.

Step 3: Pick out the four-digit and three-digit values.
Looking down the list, the only value that falls in the four-digit range, 1000 to 9999, is 4096, coming from x = 8.
The only value that falls in the three-digit range, 100 to 999, is 216, coming from x = 6.
So the largest possible four-digit value of y is 4096, and the smallest possible three-digit value of y is 216.

Step 4: Find the difference.
\[ 4096 - 216 = 3880 \]

Final Answer:
The difference between the largest four-digit value and the smallest three-digit value of y is 3880. \[ \boxed{3880} \]