Question

Let \(f\) and \(g\) be functions satisfying \[ f(x+y) = f(x)f(y), \quad f(1) = 7 \] \[ g(x+y) = g(xy), \quad g(1) = 1, \] for all \(x,y \in \mathbb{N}\). If \[ \sum_{x=1}^{n} \left(\frac{f(x)}{g(x)}\right) = 19607, \] then \(n\) is equal to

• A. \(5\)
• B. \(4\)
• C. \(6\)
• D. \(7\)

Hint:

Functional equations of the form \(f(x+y)=f(x)f(y)\) usually lead to exponential functions, while constant solutions often arise from symmetric additive relations.

Answer 1

The Correct Option is A

Step 1: Find the explicit form of \(f(x)\).
Given \[ f(x+y) = f(x)f(y), \quad f(1)=7. \] This is an exponential type functional equation. Hence, \[ f(x) = 7^x. \]

Step 2: Find the explicit form of \(g(x)\).
Given \[ g(x+y) = g(xy), \quad g(1)=1. \] Taking \(x=y=1\), \[ g(2)=g(1)=1. \] Similarly, by induction, \[ g(x)=1 \quad \text{for all } x\in\mathbb{N}. \]

Step 3: Evaluate the given sum.
\[ \sum_{x=1}^{n} \frac{f(x)}{g(x)} = \sum_{x=1}^{n} 7^x. \] This is a geometric series: \[ \sum_{x=1}^{n} 7^x = 7\left(\frac{7^n-1}{6}\right). \] Given that the sum equals \(19607\), \[ 7\left(\frac{7^n-1}{6}\right) = 19607. \] \[ 7^n - 1 = \frac{19607 \times 6}{7} = 16806. \] \[ 7^n = 16807 = 7^5. \] Thus, \[ n = 5. \]

Final Answer: \[ \boxed{5} \]