Nitin's age puzzle: last year his age was a perfect square, next year it will be a perfect cube. How many years from now until his age is again a perfect cube?

Introduction / Context:
This number puzzle blends consecutive integers with perfect squares and perfect cubes. We are told Nitin’s current age sits exactly between a square (last year) and a cube (next year). We must deduce his present age and then the waiting time until his age becomes a cube again.

Given Data / Assumptions:

• Last year: age = square of an integer.
• Next year: age = cube of an integer.
• Ages are positive integers.

Concept / Approach:
If current age is n, then n − 1 is a perfect square and n + 1 is a perfect cube. Search small squares and cubes that differ by 2. This is rare; we test candidates around typical small square–cube values.

Step-by-Step Solution:
Consider square 25 (5^2) and cube 27 (3^3): they are exactly 2 apart.Thus n − 1 = 25 and n + 1 = 27 → n = 26.Current age = 26. Last year 25 (a square), next year 27 (a cube) meets the conditions.Next time Nitin’s age is a cube: the next cube after 27 is 64 (4^3).Waiting time = 64 − 26 = 38 years.

Verification / Alternative check:
Check immediate neighbors of other small squares or cubes (e.g., 16 and 27; 36 and 27): only 25 and 27 straddle an integer. So 26 is unique in this range.

Why Other Options Are Wrong:

• 10, 27, 39, 64 years: Do not correspond to the gap between 26 and the next cube 64. 64 years is itself a cube value, not a waiting time.

Common Pitfalls:

• Assuming current age equals a square or a cube, instead of lying between them.
• Forgetting that “next year” must be the immediately next integer, hence 27 must be the cube.

Final Answer:
38 years