AP EAMCET · Maths · Binomial Theorem
For any integer \(n \geq 1\), the remainder when the expression \(n^5-5 n^3+4 n+139\) is divided by 120 is
- A 9
- B 19
- C 29
- D 39
Answer & Solution
Correct Answer
(B) 19
Step-by-step Solution
Detailed explanation
\(n^5-5 n^3+4 n+139 = (n^5-5 n^3+4 n) + 139\) \(n^5-5 n^3+4 n = n(n^4-5 n^2+4) = n(n^2-1)(n^2-4)\) \(= n(n-1)(n+1)(n-2)(n+2)\) \(= (n-2)(n-1)n(n+1)(n+2)\) This is a product of 5 consecutive integers, which is always divisible by \(5! = 120\). So,…
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