Question

A number \(1 + 7^{47}\) is divisible by \(x\). Which of the following is also divisible by \(x\)?

• A. \(7^{188} + 1\)
• B. \(7^{11} + 1\)
• C. \(7^{141} + 1\)
• D. \(7^{94} + 1\)

Hint:

Key Rule: \(x^n + a^n\) is divisible by \(x+a\) when \(n\) is odd. Look for an exponent that is an odd multiple of the original exponent (here \(141 = 3 \times 47\)).

Answer 1

The Correct Option is C

Step 1: Identify the Algebraic Form:
We are given that \(x\) divides \(7^{47} + 1\). We know the factorization identity: \(a^n + b^n\) is divisible by \(a + b\) if \(n\) is odd. Here, \(7^{141} + 1\) can be rewritten as \((7^{47})^3 + 1^3\). Step 2: Apply the Identity:
Let \(u = 7^{47}\). Then \(7^{141} + 1 = u^3 + 1\). Since 3 is odd, \(u^3 + 1\) is divisible by \(u + 1\). Therefore, \(7^{141} + 1\) is divisible by \(7^{47} + 1\). Since \(x\) divides \(7^{47} + 1\), \(x\) must also divide \(7^{141} + 1\).