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Divisibility & Remainder Test 54
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Divisibility & Remainder Test 54

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  • Question 1/10
    1 / -0
    A number when divided by divisor leaves a remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor?
    Solutions
    Short Trick:-
    When a number is divided by a divisor , if it leaves x remainder, it will leave 2x remainder when twice the number is divided
    So, according to the question, 48 should be remainder, but 11 is given
    Subtract 11 from 48 to find the divisor = 48 - 11 = 37
    Basic Method:- 
    Decoding "A number when divided by a divisor leaves a remainder of 24"

    Let the original number be 'a'.
    Let the divisor be 'd'.
    Let the quotient of dividing 'a' by 'd' be 'x'.
    Therefore, we can write the division as adad = x and the remainder is 24.
    i.e., a = dx + 24

    Decoding "When twice the original number is divided by the same divisor, the remainder is 11"

    Twice the original number is divided by d means 2a is divided by d.
    We know that a = dx + 24.
    Therefore, 2a = 2(dx + 48) or 2a = 2dx + 48

    When (2dx + 48) is divided by 'd' the remainder is 11.
    2dx is divisible by 'd' and will therefore, not leave a remainder.

    The remainder of 11 would be the remainder of dividing 48 by d.

    The question is "What number will leave a remainder of 11 when it divides 48?"
    When 37 divides 48, the remainder is 11.

    Hence, the divisor is 37.

    Choice D is the correct answer.

  • Question 2/10
    1 / -0
    Find the remainder when (3196+4197+5198) is divided by 197.
    Solutions
    According to Fermat’s theorem



    Where, p is a prime number. Now, since 197 is a prime number

  • Question 3/10
    1 / -0
    Find the remainder,
    Solutions
    According to Wilson theorem,
    If P is a prime no. then (remainder)
  • Question 4/10
    1 / -0
    By what least number should 1200 be multiplied so that it becomes a perfect square?
    Solutions
    factor of 1200 = 10 × 10 × 4 × 3
    So to make perfect square we need to multiply 1200 by 3.
    1200 × 3 = 3600 = 602
  • Question 5/10
    1 / -0
    The expression 26n - 42n, where n is a natural number is always divisible by
    Solutions
    26n - 42n = 26n – (22)2n = 26n – 24n = 24n( 22n – 1) = (24)n( 22n – 1)= (16)n( 22n – 1)
    As; 24 = 16
    Therefore (26n - 42n) is always divisible by 16.
  • Question 6/10
    1 / -0
    If 1! + 2! + 3! + 4! + .... + 100! is divided by 8, then the remainder is
    Solutions
    1! = 1 = is not divisible by 8
    2! = 2x1 = is not divisible by 8.
    3! = 3x2x1 = is not divisible by 8.
    4! = 4x3x2x1 = is divisible by 8.
    Similarly, 5!, 6!, 7!, ,,,,, 100! are all divisible by 8
    Since 4!, 5!, 6!, 7!, ,,,,, 100! all are divisible by 8, we can write it their sum as 8x
    Now ,
    1! + 2! + 3! + 8(x)
    = 1 + 2 + 6 + 8(x)
    = 8x + 8 + 1
    so, on dividing the above equation by 8,we get remainder = 1
  • Question 7/10
    1 / -0
    (320 + 321+ 322 + 323 + 324) is divisible by:
    Solutions
    (320 + 321+ 322 + 323 + 324) 



    So, the given expression is clearly divisible by 11.
  • Question 8/10
    1 / -0
    How many numbers less than 1000 are multiples of both 10 and 13?
    Solutions
    The number of multiples of 130 obtained on dividing 1000 by 130 are 7.
    Required numbers = 7
  • Question 9/10
    1 / -0
    The largest power of 4 that can divide 268! Is:
    Solutions
    The largest power of 4 that can divide 268! Is

    =[134]+[67]+[33.5]+[16.75]+[8.375]+[4.1875]+[2.09375]+[1.046875]
    = 134 + 67 + 33 + 16 + 8 + 4 + 2 + 1= 265
    So there will be total 265 power of 2 in 268! 
    The exponent of 4 will be [265/2] = 132
    Hence, 4132 will divide 268!
  • Question 10/10
    1 / -0
    What will be the largest number which when divides 37, 66 and 89, leaves a remainder of 2, 3 and 5 respectively?
    Solutions
    Let the divisor be X.
    Dividend = (Divisor × Quotient) + Reminder
    37 = (X × Q1) + 2
    (X ×Q1) = 35 [Where Q1 is a quotient]
    Similarly, 66 = (X × Q2) + 3
    (X ×Q2) = 63 [Where Q2 is a quotient]
    89 = (X × Q3) + 5
    (X ×Q3) = 84 [Where Q3 is a quotient]
    Now, since X is the largest number satisfying the given condition, X will be the HCF of 35, 63 and 84.
    HCF [35, 63, 84] = 7
    Hence, the largest number which satisfies the given requirements is 7.
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