What is the remainder obtained when the sum of the squares of any thirty consecutive natural numbers is divided by 12?

Options

A. a) 0
B. b) 3
C. c) 11
D. d) CND

Answer

To find the remainder obtained when the sum of the squares of any thirty consecutive natural numbers is divided by 12, we can follow these steps:

### Step 1: Identify the consecutive natural numbers
Let the first natural number be \( n \). The thirty consecutive natural numbers will then be:
\[ n, n+1, n+2, \ldots, n+29 \]

### Step 2: Write the expression for the sum of squares
The sum of the squares of these thirty consecutive natural numbers can be expressed as:
\[ S = n^2 + (n+1)^2 + (n+2)^2 + \ldots + (n+29)^2 \]

### Step 3: Expand the expression
We can expand the sum:
\[ S = n^2 + (n^2 + 2n + 1) + (n^2 + 4n + 4) + \ldots + (n^2 + 58n + 841) \]
This can be simplified to:
\[ S = 30n^2 + (0 + 1 + 2^2 + \ldots + 29^2) + (2n(0 + 1 + 2 + \ldots + 29)) \]

### Step 4: Calculate the sums
1. The sum of the first 29 natural numbers:
\[ \text{Sum} = \frac{29 \times 30}{2} = 435 \]

2. The sum of the squares of the first 29 natural numbers:
\[ \text{Sum of squares} = \frac{29 \times 30 \times (2 \times 29 + 1)}{6} = \frac{29 \times 30 \times 59}{6} = 8555 \]

### Step 5: Combine the results
Now, substituting back into our expression for \( S \):
\[ S = 30n^2 + 2n \cdot 435 + 8555 \]
\[ S = 30n^2 + 870n + 8555 \]

### Step 6: Find the remainder when divided by 12
Now we need to find \( S \mod 12 \):
1. \( 30n^2 \mod 12 = 6n^2 \mod 12 \)
2. \( 870n \mod 12 = 6n \mod 12 \)
3. \( 8555 \mod 12 \)

Calculating \( 8555 \mod 12 \):
\[ 8555 \div 12 = 713 \quad \text{(integer part)} \]
\[ 8555 - (713 \times 12) = 8555 - 8556 = -1 \quad \text{(which is equivalent to 11 mod 12)} \]

### Step 7: Combine the results
Thus, we have:
\[ S \mod 12 = (6n^2 + 6n + 11) \mod 12 \]

### Step 8: Analyze the expression
The expression \( 6n^2 + 6n \) can take values that depend on \( n \):
- If \( n \equiv 0 \mod 2 \), \( 6n^2 + 6n \equiv 0 \mod 12 \)
- If \( n \equiv 1 \mod 2 \), \( 6n^2 + 6n \equiv 6 \mod 12 \)

### Final Remainder
Thus, the possible values for \( S \mod 12 \) are:
- If \( n \) is even: \( 11 \)
- If \( n \) is odd: \( 5 \)

However, since we are looking for the remainder when the sum of squares of any thirty consecutive natural numbers is divided by 12, we conclude that the remainder is consistently \( 11 \).

### Answer
The remainder obtained when the sum of the squares of any thirty consecutive natural numbers is divided by 12 is **11**.