The age of Mr. X last year was the square of a number and it would be the cube of a number next year. What is the least number of years he must wait for his age to become the cube of a number again?

Correct Option (2)

The problem states that Mr. X's age last year was the square of a number, and his age next year will be the cube of a number. The difference between these two ages is two years.

Let Mr. X's age last year be represented as y² and his age next year as z³. The condition can be expressed as: z³ - y² = 2.

We systematically test perfect cubes (z³) to find a corresponding perfect square (y²):

• If z³ = 1, then y² = 1 - 2 = -1 (not a valid age).
• If z³ = 8, then y² = 8 - 2 = 6 (not a perfect square).
• If z³ = 27, then y² = 27 - 2 = 25. Since 25 is 5², this is a valid solution.

Therefore, Mr. X's age last year was 25, and his age next year will be 27. His current age is calculated as 25 + 1 = 26 years (or 27 - 1 = 26 years).

The next perfect cube following 27 is 64 (which is 4³). To determine the least number of years Mr. X must wait for his age to become a perfect cube again, we subtract his current age from the next perfect cube:

Years to wait = 64 - 26 = 38 years.

Incorrect Options:

• If Mr. X waits 42 years (Option 1), his age would be 26 + 42 = 68, which is not a perfect cube.
• If Mr. X waits 25 years (Option 3), his age would be 26 + 25 = 51, which is not a perfect cube.
• If Mr. X waits 16 years (Option 4), his age would be 26 + 16 = 42, which is not a perfect cube.

These options do not result in Mr. X's age becoming a perfect cube.