Consider the increasing sequence of natural numbers in which no term is a perfect square and no term is a perfect cube. The first few terms are 2, 3, 5, 6, 7, 10, ... What is the 300th term of this sequence?

Introduction / Context:
This problem is about constructing and analysing a special sequence of natural numbers. The sequence contains all positive integers that are neither perfect squares nor perfect cubes. We are asked to find the 300th term in this sequence. This is a classic inclusion excluding type counting problem that requires careful reasoning rather than listing all terms one by one.

Given Data / Assumptions:

• The sequence contains natural numbers greater than or equal to 1, but numbers that are perfect squares or perfect cubes are excluded.
• The first few valid terms are 2, 3, 5, 6, 7, 10, and so on, confirming that 1, 4, 8, 9, 16, 25, 27, etc. are omitted.
• We need to determine the 300th term of this sequence.
• We assume standard definitions: perfect square means n^2 for an integer n, and perfect cube means n^3 for an integer n.

Concept / Approach:
Instead of listing all numbers, we use counting. For a given positive integer N, let us count how many integers from 1 to N are excluded because they are either perfect squares or perfect cubes. The remaining numbers (from 1 to N) are exactly those that belong to the sequence. Using inclusion exclusion, we subtract the count of squares and cubes and then add back the count of sixth powers, which have been subtracted twice. Then we choose N such that the count of allowed numbers is 300 and identify the smallest such N. That N will be the 300th term of the sequence.

Step-by-Step Solution:
Step 1: For a given N, define A(N) as the number of natural numbers from 1 to N that are perfect squares.Step 2: A(N) = floor(sqrt(N)), because the squares are 1^2, 2^2, 3^2, ..., up to floor(sqrt(N))^2.Step 3: Define B(N) as the number of natural numbers from 1 to N that are perfect cubes.Step 4: B(N) = floor(N^(1/3)), because the cubes are 1^3, 2^3, 3^3, ..., up to floor(N^(1/3))^3.Step 5: Some numbers are both perfect squares and perfect cubes; these are perfect sixth powers (because lcm(2, 3) = 6). The count of such numbers from 1 to N is C(N) = floor(N^(1/6)).Step 6: The number of integers from 1 to N that are either squares or cubes is A(N) + B(N) - C(N) by inclusion exclusion.Step 7: Therefore, the number of integers from 1 to N that are neither perfect squares nor perfect cubes is S(N) = N - (A(N) + B(N) - C(N)).Step 8: We need N such that S(N) = 300.Step 9: By testing values around 320, we find that for N = 320, S(320) = 299, and for N = 321, S(321) = 300.Step 10: Hence the 300th number in the sequence is N = 321.

Verification / Alternative check:
To verify, compute A(321) = floor(sqrt(321)) = 17 (since 17^2 = 289 and 18^2 = 324 > 321). Compute B(321) = floor(321^(1/3)) = 6 (since 6^3 = 216 and 7^3 = 343 > 321). Compute C(321) = floor(321^(1/6)) = 2 (since 2^6 = 64 and 3^6 = 729 > 321). Then A(321) + B(321) - C(321) = 17 + 6 - 2 = 21. So S(321) = 321 - 21 = 300. For N = 320, the counts of squares, cubes, and sixth powers are the same, giving S(320) = 320 - 21 = 299. Thus 321 is the smallest N with 300 acceptable terms, confirming that it is the 300th term of the sequence.

Why Other Options Are Wrong:
Option 322: For N = 322, one more allowed number is added, so S(322) = 301, which would be the 301st term.Option 319: For N = 319, the count of valid numbers is less than 300, so 319 cannot be the 300th term.Option 320: As computed, S(320) = 299, so 320 corresponds to the 299th term.Option 323: This is larger than 321 and corresponds to a later term in the sequence, not the 300th.

Common Pitfalls:
Students may try to list terms manually, which is impractical for the 300th term and very error prone. Another frequent mistake is to subtract only squares or only cubes, forgetting the inclusion exclusion principle. Some also forget about sixth powers, which leads to double subtraction of numbers that are both squares and cubes. Carefully using the formula S(N) = N - (floor(sqrt(N)) + floor(N^(1/3)) - floor(N^(1/6))) avoids these mistakes.

Final Answer:
The 300th term of the sequence is 321.