6th Class Mathematics Factors and Multiples Question Bank Factors & Multiples

  • question_answer What is the least number which when divided by the numbers 3, 5, 6, 8, 10 and 12 leaves in each case a remainder of 2 but when divided by 13 leaves no remainder?

    A)  962                            

    B)  692          

    C)  269                

    D)  629

    Correct Answer: A

    Solution :

    LCM of 3, 5, 6, 8, 10 and 12 =120. So, the required number is of the form \[120\,\,k+2.\] Least value of \[k\]for which \[(120k+2)\] is divisible by 13 is\[k=8\]. \[\therefore \]Required number\[=(120\times 8+2)=962.\]

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