If n is any even number, then `n(n^(2)+20)` is always divisible by

Options

A. 15
B. 20
C. 24
D. 32

Answer

To determine if \( n(n^2 + 20) \) is always divisible by a certain number when \( n \) is any even number, we can analyze the expression step by step.

### Step 1: Define \( n \)
Let \( n \) be any even number. We can express \( n \) as \( n = 2k \), where \( k \) is an integer.

### Step 2: Substitute \( n \) into the expression
Now, substitute \( n \) into the expression \( n(n^2 + 20) \):
\[ n(n^2 + 20) = (2k)((2k)^2 + 20) \]

### Step 3: Simplify the expression
Calculate \( (2k)^2 + 20 \):
\[ (2k)^2 = 4k^2 \]
Thus,
\[ n(n^2 + 20) = 2k(4k^2 + 20) \]

### Step 4: Factor out common terms
Now, factor out the common terms:
\[ n(n^2 + 20) = 2k(4k^2 + 20) = 2k(4(k^2 + 5)) = 8k(k^2 + 5) \]

### Step 5: Analyze divisibility
From the expression \( 8k(k^2 + 5) \), we can see that:
- \( 8k \) is clearly divisible by \( 8 \) since \( k \) is an integer.
- The term \( k^2 + 5 \) is an integer, but we need to check if \( 8k(k^2 + 5) \) has any further divisibility.

### Step 6: Check specific even values of \( n \)
Let's check with specific even values of \( n \):
1. For \( n = 2 \):
\[ 2(2^2 + 20) = 2(4 + 20) = 2 \times 24 = 48 \]
\( 48 \) is divisible by \( 24 \).

2. For \( n = 4 \):
\[ 4(4^2 + 20) = 4(16 + 20) = 4 \times 36 = 144 \]
\( 144 \) is also divisible by \( 24 \).

3. For \( n = 6 \):
\[ 6(6^2 + 20) = 6(36 + 20) = 6 \times 56 = 336 \]
\( 336 \) is also divisible by \( 24 \).

### Conclusion
From the calculations, we can conclude that \( n(n^2 + 20) \) is always divisible by \( 24 \) for any even number \( n \).

### Final Answer
Thus, \( n(n^2 + 20) \) is always divisible by \( 24 \).

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