The problem in the image states:

Q: The sum of six consecutive positive odd integers starting with $n$ is a perfect cube. Find the smallest possible $n$.

Let the six consecutive odd integers be: $n, n+2, n+4, n+6, n+8, n+10.$

The sum of these integers is: $S = n + (n+2) + (n+4) + (n+6) + (n+8) + (n+10).$ Simplifying this: $S = 6n + (2 + 4 + 6 + 8 + 10) = 6n + 30.$ We are told that this sum is a perfect cube: $6n + 30 = k^3 \text{ for some integer } k.$ Thus, we have: $6n = k^3 - 30.$ So: $n = \frac{k^3 - 30}{6}.$ To find the smallest possible $n$, we need $k^3 - 30$ to be divisible by 6. Let's test small values of $k$.

1. For $k = 3$: $k^3 = 3^3 = 27,$ $6n = 27 - 30 = -3 \quad \text{(not divisible by 6)}.$ So $k = 3$ doesn't work.

2. For $k = 4$: $k^3 = 4^3 = 64,$ $6n = 64 - 30 = 34 \quad \text{(not divisible by 6)}.$ So $k = 4$ doesn't work.

3. For $k = 5$: $k^3 = 5^3 = 125,$ $6n = 125 - 30 = 95 \quad \text{(not divisible by 6)}.$ So $k = 5$ doesn't work.

4. For $k = 6$: $k^3 = 6^3 = 216,$ $6n = 216 - 30 = 186 \quad \text{(divisible by 6)}.$ Thus, $n = \frac{186}{6} = 31$.

Therefore, the smallest possible value of $n$ is 31.

The correct answer is d) 31.

Would you like further details or have any questions about this solution?

Here are some related questions you might find interesting:

1. How do we find the sum of consecutive odd integers in general?
2. Why is it important to check if $k^3 - 30$ is divisible by 6?
3. Can we generalize this approach to solve similar problems for different sequences of odd integers?
4. What are the properties of perfect cubes in number theory?
5. How would the problem change if we considered even integers instead of odd ones?

Tip: When working with consecutive integers, their sum often forms a recognizable pattern, making it easier to solve problems involving sums of sequences.