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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Mathematics
Chapter
Basics of Mathematics

BITSAT 2023 · Paper 2 · Mathematics · Basics of Mathematics

If n is a positive integer, then 2.4^ 2 n+1 +3^ 3 n+1 is divisible...

If nn is a positive integer, then 2.42n+1+33n+12.4^{2 n+1}+3^{3 n+1} is divisible by:

Options

  1. A
    2
  2. B
    7
  3. C
    11
  4. D
    27

Correct answer

C.
11

Step-by-step solution

Let P(n)=2⋅42n+1+33n+1P(n)=2 \cdot 4^{2 n+1}+3^{3 n+1}
Then P(1)=2.43+34=209P(1)=2.4^3+3^4=209, which is divisible by 11 but not divisible by 2,7 or 27.
Further, let P(k)=2.42k+1+33k+1P(k)=2.4^{2 k+1}+3^{3 k+1} is divisible by 11 , that is,
2.42k+1+33k+1=11q2.4^{2 k+1}+3^{3 k+1}=11 q for some integer q. Now
P(k+1)=2⋅42k+3+33k+4=2⋅42k+1⋅42+33k+1⋅33=16⋅2⋅42k+1+27⋅33k+1=16⋅2⋅42k+1+(16+11)⋅33k+1=16[2⋅42k+1+33k+1]+11⋅33k+1=16⋅11q+11⋅33k+1=11(16q+33k+1)=11m\begin{aligned} \\ \mathrm{P}(\mathrm{k}+1) & =2 \cdot 4^{2 \mathrm{k}+3}+3^{3 \mathrm{k}+4} \\ \\ & =2 \cdot 4^{2 \mathrm{k}+1} \cdot 4^2+3^{3 \mathrm{k}+1} \cdot 3^3 \\ \\ & =16 \cdot 2 \cdot 4^{2 \mathrm{k}+1}+27 \cdot 3^{3 \mathrm{k}+1} \\ \\ & =16 \cdot 2 \cdot 4^{2 \mathrm{k}+1}+(16+11) \cdot 3^{3 \mathrm{k}+1} \\ \\ & =16\left[2 \cdot 4^{2 \mathrm{k}+1}+3^{3 \mathrm{k}+1}\right]+11 \cdot 3^{3 \mathrm{k}+1} \\ \\ & =16 \cdot 11 \mathrm{q}+11 \cdot 3^{3 \mathrm{k}+1} \\ \\ & =11\left(16 \mathrm{q}+3^{3 \mathrm{k}+1}\right)=11 \mathrm{ m} \\ \end{aligned}
where m=16q+33k+1m=16 q+3^{3 k+1} is another integer.
∴P(k+1)\therefore \mathrm{P}(\mathrm{k}+1) is divisible by 11.
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