COMEDK · Maths · 1. Basic of Mathematics
\(2^{3 n}-7 n-1\) is divisible by
- A 64
- B 36
- C 49
- D 25
Answer & Solution
Correct Answer
(C) 49
Step-by-step Solution
Detailed explanation
Let \(f(n) = 2^{3n} - 7n - 1 = 8^n - 7n - 1\). For \(n = 1\), \(f(1) = 8^1 - 7(1) - 1 = 8 - 8 = 0\), which is divisible by any integer. For \(n = 2\), \(f(2) = 8^2 - 7(2) - 1 = 64 - 14 - 1 = 49\). For \(n = 3\), \(f(3) = 8^3 - 7(3) - 1 = 512 - 21 - 1 = 490\). Since \(f(2) = 49\)…
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