Question

How many numbers between 1 and 1000 (both excluded) are both squares and cubes?

• A. None
• B. 1
• C. 2
• D. 3

Hint:

A number that is both a square and a cube must be a perfect sixth power, since 6 is the lowest common multiple of 2 and 3; check which sixth powers fall strictly between 1 and 1000.

Answer 1

The Correct Option is B

Step 1: Understand what it means for a number to be both a square and a cube.
A perfect square can be written as \(n^2\) for some whole number \(n\), and a perfect cube can be written as \(m^3\) for some whole number \(m\). A number that is both a square and a cube at once must be a perfect sixth power, that is, of the form \(k^6\) for some whole number \(k\), because 6 is the smallest power that is a multiple of both 2 and 3.

Step 2: List the perfect sixth powers near the given range.
\[ 1^6 = 1, \quad 2^6 = 64, \quad 3^6 = 729, \quad 4^6 = 4096 \]
We stop at \(4^6\) since it already exceeds 1000.

Step 3: Apply the condition that both 1 and 1000 are excluded.
The question asks for numbers strictly between 1 and 1000. \(1^6 = 1\) does not count, since it equals the excluded lower bound, and \(4^6 = 4096\) does not count either, since it is far above 1000. This leaves \(2^6 = 64\) and \(3^6 = 729\), both of which lie strictly between 1 and 1000.

Step 4: Verify both values directly.
\(64 = 8^2\) and also \(64 = 4^3\), so 64 is genuinely both a square and a cube.
\(729 = 27^2\) and also \(729 = 9^3\), so 729 is genuinely both a square and a cube.

Step 5: Count the qualifying numbers and check this against the printed answer key.
By direct count, there are 2 such numbers, 64 and 729, strictly between 1 and 1000, which corresponds to option (C). The answer key printed with this paper marks option (B) as correct for this question. Working through the sixth-power condition carefully, and checking every candidate by hand, does not reproduce that value: this looks like an error carried over from the source answer key rather than a gap in the method shown here. This discrepancy should be flagged for review rather than silently accepted.

Final Answer:
Direct counting of perfect sixth powers strictly between 1 and 1000 gives 2 qualifying numbers, 64 and 729. \[ \boxed{2} \]