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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Basics of Mathematics

NDA 2018 · Mathematics · Basics of Mathematics

If n ∈ N , then 121^ n -25^ n +1900^ n -(-4)^ n is divisible by...

If n∈N\mathrm{n} \in \mathrm{N}, then 121n−25n+1900n−(−4)n121^{\mathrm{n}}-25^{\mathrm{n}}+1900^{\mathrm{n}}-(-4)^{\mathrm{n}} is divisible by which one of the following?

Options

  1. A19041904
  2. B20002000
  3. C20022002
  4. D20062006

Correct answer

B. 20002000

Step-by-step solution

n∈N121n−25n+1900n−(−4n)\begin{aligned} & \mathrm{n} \in \mathrm{ N} \\ & 121^{\mathrm{n}}-25^{\mathrm{n}}+1900^{\mathrm{n}}-\left(-4^{\mathrm{n}}\right) \end{aligned} Let us substitute n=1\mathrm{n}=1 We get, (121)1−(25)1+(1900)1−(−41)=121−25+1900+4=2025−25=2000(121)^1-(25)^1+(1900)^1-\left(-4^1\right)=121-25+1900+4=2025-25=2000 So, given expression is divisible by 2000
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