Question

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

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

Hint:

For expressions like \( (1+k)^n \): \begin{itemize} \item Use binomial expansion modulo powers. \item Higher terms vanish modulo \( k^3 \). \end{itemize}

Answer 1

The Correct Option is A

Concept: We test divisibility using modular arithmetic. Observe: \[ 2^{4n} = (16)^n \] Work modulo \( 125 = 5^3 \). Step 1: {\color{red}Check modulo 5.} \[ 16 \equiv 1 \pmod{5} \Rightarrow 16^n \equiv 1 \] So: \[ 2^{4n} - 15n - 1 \equiv 1 - 0 - 1 = 0 \pmod{5} \] Step 2: {\color{red}Check modulo 25.} \[ 16 \equiv -9 \pmod{25} \] Use binomial pattern: \[ 16^n = (1+15)^n \equiv 1 + 15n \pmod{25} \] So: \[ 2^{4n} - 15n - 1 \equiv (1+15n) - 15n - 1 = 0 \] Step 3: {\color{red}Check modulo 125.} Similarly expand: \[ 16 = 1 + 15 \] Using binomial expansion: \[ (1+15)^n = 1 + 15n + \frac{n(n-1)}{2}15^2 + \cdots \] Higher terms multiples of 125 vanish mod 125. So: \[ 16^n \equiv 1 + 15n \pmod{125} \] Hence: \[ 2^{4n} - 15n - 1 \equiv 0 \pmod{125} \] Step 4: {\color{red}Conclusion.} Divisible by \( 125 \).