Consider the sequence of natural numbers that excludes all perfect squares and all perfect cubes. The beginning of this sequence is 2, 3, 5, 6, 10, .... What is the 300th term of this sequence?

Introduction:
This problem focuses on number theory and counting techniques. The sequence consists of natural numbers that are neither perfect squares nor perfect cubes. Your task is to determine which number appears at the 300th position in this specially filtered list of natural numbers.

Given Data / Assumptions:

• The sequence includes natural numbers only.
• Every perfect square (like 1, 4, 9, 16, ...) is excluded.
• Every perfect cube (like 1, 8, 27, ...) is excluded.
• If a number is both a perfect square and a perfect cube, that is, a perfect sixth power (like 1, 64, 729), it is also excluded.
• We need to find the 300th term in this filtered sequence.

Concept / Approach:
For a given positive integer N, let us count how many numbers from 1 to N remain after removing all perfect squares and perfect cubes, while avoiding double subtraction of perfect sixth powers. The count of valid terms up to N is:
valid(N) = N - (number of perfect squares up to N) - (number of perfect cubes up to N) + (number of perfect sixth powers up to N). We then search for N such that valid(N) = 300. That N will be the 300th term of the sequence.

Step-by-Step Solution:
Step 1: Let N be the 300th term, so valid(N) = 300. Step 2: For a trial value N = 321, compute: - Number of perfect squares ≤ 321 is floor(sqrt(321)) = 17. - Number of perfect cubes ≤ 321 is floor(cuberoot(321)) = 6. Step 3: Numbers that are both squares and cubes are perfect sixth powers. The largest sixth power ≤ 321 is 2^6 = 64, so there are 2 such numbers (1 and 64). Step 4: Apply the formula: valid(321) = 321 - 17 - 6 + 2 = 321 - 21 = 300. Step 5: Therefore, the 300th number in the sequence is 321.

Verification / Alternative check:
We can check a nearby value. For N = 320: number of squares = 17, cubes = 6, sixth powers = 2, so valid(320) = 320 - 17 - 6 + 2 = 299. Thus the 299th term is 320 and the very next term, the 300th, is 321. This confirms our result.

Why Other Options Are Wrong:

• 319 and 320: They correspond to earlier positions (298th or 299th), not the 300th term.
• 322 and 318: These would give counts greater or less than 300 when substituted into the counting formula.

Common Pitfalls:
The main difficulty is double counting numbers that are both perfect squares and cubes. If you subtract both squares and cubes without adding back the intersection (perfect sixth powers), you undercount valid numbers. It is also easy to miscalculate the integer parts of square roots or cube roots. Careful computation and checking adjacent values ensures accuracy.

Final Answer:
The 300th term of the sequence of natural numbers that are neither squares nor cubes is 321.