If A is the sum of the squares of the first n natural numbers (where n < 100), then for how many values of n will A be divisible by 5?

Options

A. 40
B. 60
C. 59
D. 39

Answer

To solve the problem of how many values of \( n \) (where \( n < 100 \)) make the sum of the squares of the first \( n \) natural numbers divisible by 5, we can follow these steps:

### Step 1: Understand the formula for the sum of squares
The sum of the squares of the first \( n \) natural numbers is given by the formula:
\[ A = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 2: Determine the condition for divisibility by 5
We need to find when \( A \) is divisible by 5. This means we need to check when:
\[ \frac{n(n + 1)(2n + 1)}{6} \equiv 0 \mod 5 \]
This can be simplified to checking when:
\[ n(n + 1)(2n + 1) \equiv 0 \mod 30 \]
since \( 6 \) and \( 5 \) are coprime.

### Step 3: Analyze \( n(n + 1)(2n + 1) \mod 5 \)
We will check the values of \( n \) from 1 to 99 and see when \( n(n + 1)(2n + 1) \) is divisible by 5.

- \( n \equiv 0 \mod 5 \): Here \( n \) is divisible by 5.
- \( n \equiv 4 \mod 5 \): Here \( n + 1 \equiv 0 \mod 5 \).
- \( n \equiv 2 \mod 5 \): Here \( 2n + 1 \equiv 0 \mod 5 \).

### Step 4: Count the valid \( n \)
Now we will count how many values of \( n \) satisfy these conditions:

1. **For \( n \equiv 0 \mod 5 \)**:
- The values are: 5, 10, 15, ..., 95.
- This is an arithmetic sequence where \( a = 5 \), \( d = 5 \), and the last term is 95.
- The number of terms is given by:
\[ n = \frac{95 - 5}{5} + 1 = 19 \]

2. **For \( n \equiv 4 \mod 5 \)**:
- The values are: 4, 9, 14, ..., 94.
- This is also an arithmetic sequence where \( a = 4 \), \( d = 5 \), and the last term is 94.
- The number of terms is:
\[ n = \frac{94 - 4}{5} + 1 = 19 \]

3. **For \( n \equiv 2 \mod 5 \)**:
- The values are: 2, 7, 12, ..., 92.
- This is an arithmetic sequence where \( a = 2 \), \( d = 5 \), and the last term is 92.
- The number of terms is:
\[ n = \frac{92 - 2}{5} + 1 = 19 \]

### Step 5: Sum the counts
Now we sum the counts from all three cases:
\[ 19 + 19 + 19 = 57 \]

### Conclusion
Thus, the total number of values of \( n \) for which \( A \) is divisible by 5 is **57**.

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