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
• E. 4280

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.

Answer 1

The Correct Option is D

Step 1: Understand when y is a valid (integer) value.
\(y=x^{x/2}=\left(\sqrt{x}\right)^x\). For \(y\) to actually be a whole-number "value" (a four-digit or three-digit number, as asked), \(x^{x/2}\) must be an integer. This happens in two cases: (i) \(x\) is even, so the exponent \(x/2\) is a whole number and \(x^{x/2}\) is automatically an integer; or (ii) \(x\) is odd but \(x\) is a perfect square, so \(\sqrt{x}\) is itself a whole number and \(y=(\sqrt{x})^x\) is an integer. If \(x\) is odd and not a perfect square, \(y\) is irrational and has no "digit value", so such \(x\) is excluded.

Step 2: Compute y for small values of x.
\(x=1\) (odd, perfect square): \(y=1^{0.5}=1\) (1-digit)
\(x=2\) (even): \(y=2^{1}=2\) (1-digit)
\(x=3\) (odd, not a perfect square): excluded (irrational)
\(x=4\) (even): \(y=4^{2}=16\) (2-digit)
\(x=5\) (odd, not a perfect square): excluded
\(x=6\) (even): \(y=6^{3}=216\) (3-digit)
\(x=7\) (odd, not a perfect square): excluded
\(x=8\) (even): \(y=8^{4}=4096\) (4-digit)
\(x=9\) (odd, perfect square): \(y=9^{4.5}=9^4\times\sqrt9=6561\times3=19,683\) (5-digit)
\(x=10\) (even): \(y=10^{5}=1,00,000\) (6-digit)

Step 3: Identify the smallest three-digit value of y.
Scanning the list, the only value of \(y\) that falls in the three-digit range (100 to 999) is \(216\), coming from \(x=6\). Since it is the only three-digit value produced, it is automatically also the smallest one: smallest three-digit \(y=216\).

Step 4: Identify the largest four-digit value of y.
Scanning further, the only value of \(y\) that falls in the four-digit range (1000 to 9999) is \(4096\), coming from \(x=8\) (the next valid \(x=9\) already jumps to a 5-digit value, \(19,683\)). So it is also the only, hence the largest, four-digit value: largest four-digit \(y=4096\).

Step 5: Find the required difference.
$$4096-216=3880$$
The correct option is (d) 3880.