The sum of thirty-two consecutive natural numbers is a perfect square. What is the least possible sum of the smallest and the largest of the thirty-two numbers?

Options

A. 81
B. 35
C. 49
D. 64

Answer

To solve the problem of finding the least possible sum of the smallest and largest of thirty-two consecutive natural numbers such that their sum is a perfect square, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Consecutive Natural Numbers:**
Let the smallest of the 32 consecutive natural numbers be \( a \). Therefore, the numbers can be represented as:
\[ a, a+1, a+2, \ldots, a+31 \]

2. **Calculate the Sum of the Numbers:**
The sum \( S \) of these 32 numbers can be calculated using the formula for the sum of an arithmetic series:
\[ S = \frac{n}{2} \times (2a + (n-1)d) \]
Here, \( n = 32 \) and \( d = 1 \) (the common difference). Thus, the formula simplifies to:
\[ S = \frac{32}{2} \times (2a + 31) = 16 \times (2a + 31) = 32a + 496 \]

3. **Set the Condition for a Perfect Square:**
We need \( S = 32a + 496 \) to be a perfect square. Let’s denote it as:
\[ S = k^2 \quad \text{for some integer } k \]
Therefore, we have:
\[ 32a + 496 = k^2 \]
Rearranging gives:
\[ 32a = k^2 - 496 \]
Hence,
\[ a = \frac{k^2 - 496}{32} \]

4. **Finding Suitable Values for \( k \):**
For \( a \) to be a natural number, \( k^2 - 496 \) must be divisible by 32. We can check values of \( k \) such that \( k^2 \) is greater than 496 and \( k^2 - 496 \equiv 0 \mod 32 \).

5. **Testing Values of \( k \):**
We can start testing values of \( k \):
- For \( k = 28 \):
\[ k^2 = 784 \quad \Rightarrow \quad 784 - 496 = 288 \quad \Rightarrow \quad a = \frac{288}{32} = 9 \]
- For \( k = 29 \):
\[ k^2 = 841 \quad \Rightarrow \quad 841 - 496 = 345 \quad \Rightarrow \quad a = \frac{345}{32} \text{ (not an integer)} \]
- For \( k = 30 \):
\[ k^2 = 900 \quad \Rightarrow \quad 900 - 496 = 404 \quad \Rightarrow \quad a = \frac{404}{32} \text{ (not an integer)} \]
- Continuing this process, we find that \( k = 28 \) gives us a valid \( a \).

6. **Calculate the Largest Number:**
The largest number in the series is:
\[ a + 31 = 9 + 31 = 40 \]

7. **Calculate the Sum of the Smallest and Largest:**
Finally, the sum of the smallest and largest numbers is:
\[ a + (a + 31) = 9 + 40 = 49 \]

### Final Answer:
The least possible sum of the smallest and the largest of the thirty-two numbers is **49**.