Question

The expression \( 2^{4n} - 15n - 1 \), where \( n \in \mathbb{N} \) (the set of natural numbers), is divisible by:

• A. \( 125 \)
• B. \( 225 \)
• C. \( 325 \)
• D. \( 425 \)

Hint:

When you encounter divisibility questions involving powers, test small values of \( n \) to find a consistent pattern, especially when the expression simplifies nicely using powers of 2, 4, or 16.

Answer 1

The Correct Option is B

Step 1: Rewrite the given expression. Given expression: \[ 2^{4n} - 15n - 1 \] Since: \[ 2^{4n} = (2^4)^n = 16^n \] the expression becomes: \[ 16^n - 15n - 1 \]
Step 2: Check with small values of \( n \). For \( n = 1 \): \[ 16^1 - 15(1) - 1 = 16 - 15 - 1 = 0 \quad \text{(divisible by any number)} \] For \( n = 2 \): \[ 16^2 - 15(2) - 1 = 256 - 30 - 1 = 225 \] and \[ 225 \div 225 = 1 \quad \text{(exact division)} \] For \( n = 3 \): \[ 16^3 - 15(3) - 1 = 4096 - 45 - 1 = 4050 \] and \[ 4050 \div 225 = 18 \quad \text{(exact division)} \] Thus, the expression is divisible by \( 225 \) for multiple values of \( n \).
Step 3: Conclude the answer. Therefore, \( 2^{4n} - 15n - 1 \) is divisible by \( 225 \).