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CDS - Divisibility Rule Test 967
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CDS - Divisibility Rule Test 967
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  • Question 1/5
    1 / -0.33
    When 1062, 1134 and 1182 are divided by the greatest number x, the remainder in each case is y. What is the value of (x – y)?
    Solutions

    Difference between numbers

    1134 – 1062 = 72 and 1182 – 1134 = 48

    Greatest Number (that leaves same remainder) = HCF of (72, 48) = 24

    Therefore, x = 24

    And remainder (y) =  = 6

    Therefore, x – y = 24 – 6 = 18

  • Question 2/5
    1 / -0.33
    Consider the following assumption and two statements :
    Assumption : A number ‘ABCDE’ is divisible by 11.
    Statement-I : E – D + C – B + A is divisible by 11.
    Statement- II : E - D + C – B + A =0.
    Which one of the following is correct ?
    Solutions
    We know that, if the difference of the sum of odd digits and sum of even digits is either 0 or multiple of 11, then the number is divisible by 11
    Given number is ABCDE
    Her A+C+E-(B+D)=0 or divisible by 11
    Hence both statements are true.
  • Question 3/5
    1 / -0.33
     is

    I: divisible by 24

    II: not divisible by 24

    III: divisible by 10

    Solutions

    We know that  is divisible by , only when  is even.

    So,  is not divisible by 24, i.e. ( 17+7 ) , power is not even.

      is divisible by , for all values of  i.e.  = odd or even

     is divisible by  i.e. .

  • Question 4/5
    1 / -0.33
    A 95-digit number is formed by writing the first x natural numbers in front of each other as 12345678910111213… Find the remainder when this number is divided by 16.
    Solutions

    We know that When we divide a number (greater than four digits) by 16, we can obtain remainder simply by dividing last four digit of that number.

    Now total digits = 95

    One digit natural number = 9

    Remaining digits = 95 – 9 =86

    After that we will write two digit natural number.

    Now 43 2 = 86

    So we can understand that we have to write 43 two-digit natural number (10-52).

    Now, Last three digits of the resultant number = 5152

    When we divide 5152 by 16, we obtain 0 as a remainder

    So required remainder = 0

  • Question 5/5
    1 / -0.33
    Find the least number which should be added to 2497 so that sum is exactly divisible by 5,6,4 and 3.
    Solutions

    We have,

    LCM ( 5, 6, 4, 3) =

    = 60

    On dividing 2497 by 60, remainder = 37

     Least number to be added = 60 – 37 = 23.

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