Number System


  1. The numbers 2272 and 875 are divided by a 3-digit number N, giving the same remainders. The sum of the digits of N is









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    Let the remainder in each case be x.
    Then, (2272 – x) and (875 – x) are exactly divisible by that three digit number.
    Hence, their difference [(2272 – x) – (875 – x)] = 1397 will also be exactly divisible by the said divisor (N).
    Now, 1397 = 11×127 Since both 11 and 127 are prime numbers, N is 127.
    ∴  Sum of digits = 1+ 2 + 7 = 10

    Correct Option: A

    Let the remainder in each case be x.
    Then, (2272 – x) and (875 – x) are exactly divisible by that three digit number.
    Hence, their difference [(2272 – x) – (875 – x)] = 1397 will also be exactly divisible by the said divisor (N).
    Now, 1397 = 11×127 Since both 11 and 127 are prime numbers, N is 127.
    ∴  Sum of digits = 1+ 2 + 7 = 10


  1. A 2-digit number is 3 times the sum of its digits. If 45 is added to the number, its digits are interchanged. The sum of digits of the number is









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    Let the two digit number be = 10x + y.
    According to the question,
    10 x + y = 3(x + y)
    ⇒  10 x + y = 3x + 3y
    ⇒  10 x + y – 3x –3y = 0
    ⇒  7 x – 2y = 0    ....(i)
    and,
    10 x + y + 45= 10 y + x
    ⇒  10 y + x – 10x –y = 45
    ⇒  9 y – 9 x = 45
    ⇒  9 (y – x) = 45
    ⇒  y – x = 5    ...(ii)
    2 × (ii) + (i) we have
    2y –2x + 7x – 2y =10

    ⇒  5x = 10 ⇒ x =
    10
    = 2
    5

    From equation (ii),
    y –2 = 5 ⇒ y = 2 + 5 = 7
    ∴  Number = 10x + y
    = 2 × 10 +7 = 27
    ∴  Sum of digits = 2 + 7 = 9

    Correct Option: B

    Let the two digit number be = 10x + y.
    According to the question,
    10 x + y = 3(x + y)
    ⇒  10 x + y = 3x + 3y
    ⇒  10 x + y – 3x –3y = 0
    ⇒  7 x – 2y = 0    ....(i)
    and,
    10 x + y + 45= 10 y + x
    ⇒  10 y + x – 10x –y = 45
    ⇒  9 y – 9 x = 45
    ⇒  9 (y – x) = 45
    ⇒  y – x = 5    ...(ii)
    2 × (ii) + (i) we have
    2y –2x + 7x – 2y =10

    ⇒  5x = 10 ⇒ x =
    10
    = 2
    5

    From equation (ii),
    y –2 = 5 ⇒ y = 2 + 5 = 7
    ∴  Number = 10x + y
    = 2 × 10 +7 = 27
    ∴  Sum of digits = 2 + 7 = 9



  1. On multiplying a number by 7, all the digits in the product appear as 3’s. the smallest such number is









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    Let the smallest number be x.
    ∴  x × 7 = 33333....

    ⇒  x =
    33333....
    = 47619
    7

    Correct Option: C

    Let the smallest number be x.
    ∴  x × 7 = 33333....

    ⇒  x =
    33333....
    = 47619
    7


  1. 7 is added to a certain number; the sum is multiplied by 5; the product is divided by 9 and 3 is subtracted from the quotient. Thus if the remainder left is 12, what was the original number ?









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    Check through options
    20 → 20 + 7 = 27 → 27 × 5
    = 135 → 135 ÷ 9
    = 15 → 15 – 3 = 12
    OR, We will solve the problem from the opposite side.
    Here the remainder is 12.
    12 + 3 = 15
    15 × 9 = 135
    135 ÷ 5 = 27
    27 – 7 = 20
    ∴  The original number was 20.

    Correct Option: B

    Check through options
    20 → 20 + 7 = 27 → 27 × 5
    = 135 → 135 ÷ 9
    = 15 → 15 – 3 = 12
    OR, We will solve the problem from the opposite side.
    Here the remainder is 12.
    12 + 3 = 15
    15 × 9 = 135
    135 ÷ 5 = 27
    27 – 7 = 20
    ∴  The original number was 20.



  1. If the difference of two numbers is 3 and the difference of their squares is 39, then the larger number is









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    Let the numbers be a and b,
    where a > b
    According to the question,
    a – b = 3    ....(i)
    a2 − b2 = 39
    ⇒  (a + b) (a – b) = 39

    ⇒  a + b =
    39
    =
    39
    = 13
    a − b3

    ⇒  a + b = 3    .....(ii)
    Adding equations (i) and (ii)
    2a = 16 ⇒ a = 8

    Correct Option: A

    Let the numbers be a and b,
    where a > b
    According to the question,
    a – b = 3    ....(i)
    a2 − b2 = 39
    ⇒  (a + b) (a – b) = 39

    ⇒  a + b =
    39
    =
    39
    = 13
    a − b3

    ⇒  a + b = 3    .....(ii)
    Adding equations (i) and (ii)
    2a = 16 ⇒ a = 8