2. Solved Papers
  3. CET Karnataka Engineering

CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2010

  • question_answer
    The greatest value of x satisfying \[~21=385\](mod x) and \[587=167\] (mod x) is

    A) \[156\]                                 

    B) \[32\]

    C) \[28\]                                   

    D) \[56\]

    Correct Answer: C

    Solution :

     We know that, \[a\equiv b(\bmod x)=\frac{(a-b)}{x}\]                 Given,  \[21\equiv 385(\bmod x)=\frac{(21-385)}{x}\]                                 \[=-\frac{364}{x}\]           ….(i)                 and \[587\equiv 167(\bmod x)\]                 \[=\frac{(587-167)}{x}=\frac{420}{x}\]    …..(ii) Now, the greatest value of ‘x’ satisfying Eq. (i) and Eq. (ii) \[=\max [LCM\,of\,(364,\,420)]\] \[\Rightarrow \]               \[x=\max \,(13,\,15,\,28)\] \[\Rightarrow \]               \[x=28\]


You need to login to perform this action.
You will be redirected in 3 sec spinner