Q42 / 90 · KVPY 2010 (SB/SX) · Mathematics

Let $a_{0}=0$ and $a_{x}=3 a_{n-1}+1$ for $n \geq 1$. Then the remainder obtained dividing aco ob by 11 is-

  • Correct answer
    A
    0
  • B
    7
  • C
    3
  • D
    4

Solution

$$ \begin{array}{l} a_{n}=3 a_{n-1}+1 \\ a_{2010}=3 a_{2009}+1 \\ =3\left(3 a_{2008}+1\right)+1=3^{2} a_{2008}+3+1 \\ \quad=3^{3} a_{2007}+3+3+1 \end{array} $$ $$ \begin{array}{l} 3^{2010} \mathrm{a}_{0}+\underbrace{(3+3+\ldots 3)}_{2009 \text { times }}+1 \\ =0+6027+1=6028 \end{array} $$ Remainder $\left(\frac{6028}{11}\right)=0$

Hence, the correct answer is option: 'A'.

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