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Theory of Numbers Test - 7

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Theory of Numbers Test - 7
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  • Question 1
    1 / -0
    Suppose x and y are the digits in the number x01y such that it is divisible by both 8 and 9. What is the sum of x and y?
    Solution
    Given number x01y is divisible by both 8 and 9.
    For divisibility by 8: The number formed by the last three digits should be evenly divisible by 8.
    For divisibility by 9: The sum of all the digits should be divisible by 9.
    x + y + 1 = 9, and y can only be 6 (as 016 is divisible by 8).

    x + 6 + 1 = 9 or Multiple of 9
    x + 7 = 9
    x has to be 2.
    So, x + y = 2 + 6 = 8
  • Question 2
    1 / -0
    56 - 1 is divisible by
    Solution
    (56 - 1) = (53)2 - (1)2
    = (125)2 - (1)2 = (125 + 1) (125 - 1)
    = 126 × 124
    = 31 × 4 × 126
    Therefore, it is clear that the expression is divisible by 31.
  • Question 3
    1 / -0
    How many pairs of x and y are possible in the eight-digit number x456789y, if the number is divisible by 18?
    Solution
    For the number to be divisible by 18, it should be divisible by both 2 and 9.
    Now, sum of the digits of the number = x + y + 39
    So, x + y should be 6 or 15.
    Next value should be 24, but since x and y are distinct digits, it is not possible.
    The possible pairs of (x, y) are (0, 6), (6, 0), (1, 5), (5, 1), (2, 4), (4, 2), (7, 8), (8, 7), (9, 6) (6, 9) and (3, 3).
    Since the number has to be divisible by 2, the last digit should be even.
    Then, there are 6 favourable cases. But (0, 6) is not possible, as then the given number will not be an eight-digit number.
    Hence, the number of possible pairs is 5.
  • Question 4
    1 / -0
    A number formed by writing any digit 6 times is always divisible by
    Solution
    When a digit is written six times together, i.e. aaaaaa,
    Then factors of aaaaaa is = 1001 × (aaa).
    or, aaaaaa = 1001 × (aaa) = 7 × 11 × 13 × (aaa)
    Hence, the given number is divisible by 7, 11, 13 and 1001.
  • Question 5
    1 / -0
    What will be the remainder when (1234 × 1235 × 1236) is divided by 11?
    Solution
    The remainder when 1234 × 1235 × 1236 is divided by 11, is
    or or or 2.
  • Question 6
    1 / -0
    A 101-digit number is formed by writing natural numbers (starting from 1) as 12345678910111213…. Find the remainder when this number is divided by 8.
    Solution
    The first 11 digits of the number are from 1 to 10.
    Now we have another 90 digits to write.
    Each of these 90 digits will come in pairs from a 2-digit number.
    Thus, 45 numbers more will be required as 45 × 2 = 90
    Now, the last number to be filled in will be 10 + 45 = 55
    Before this, 54 shall be filled in.
    Thus, the last 3 digits of the number will be 455. Thus, the number so obtained will be of the form 1000y + 455.
    Now, 1000y = 8a
    Thus, the only remainder term will come from 455.
    Now, 455 = 56 × 8 + 7
    Therefore, remainder = 7
  • Question 7
    1 / -0
    The HCF of two numbers is 21 and their sum is 105. The LCM of the numbers is
    Solution
    Let the two numbers be 21x and 21y.
    21x + 21y = 105
    x + y = 5 where x, y are co-prime numbers
    x = 1, y = 4 or x = 2, y = 3
    Numbers = (21 × 1 and 21 × 4) or (21 × 2 and 21 × 3) = (21 and 84) or (42 and 63)
    LCM of 21 and 84 is 84 and LCM of 42 and 63 is 126.
  • Question 8
    1 / -0
    There are 4 numbers. The HCF of each pair is 3 and the LCM of all the 4 numbers is 126. What is the product of the 4 numbers?
    Solution
    Product of 'n' numbers = [(HCF for each pair)n-1 × (LCM of 'n' numbers)]
    = 33 × 126 = 27 × 126 = 3,402
  • Question 9
    1 / -0
    If x136y is a five digit number divisible by 15, where x and y are digits, then the least possible value of x + y is
    Solution
    x136y is a five digit number divisible by 15, means number should be divisible by both 3 and 5.
    Number divisible by 5 means y should be 0 or 5.

    Case 1: y = 0
    x1360 is divisible by 3, so sum of digits should be multiple of 3.
    Sum of digits = 10 + x
    Possible value of x = 2, 5 and 8.

    Case 2: y = 5
    x1365 is divisible by 3, so sum of digits should be multiple of 3.
    Sum of digits = 15 + x
    Possible value of x = 0, 3, 6 and 9.

    For least value of x + y, take y = 0 and x = 2
    So, x + y = 2
  • Question 10
    1 / -0
    If 7A28 is divisible by 8, where A is a digit, then how many such numbers are possible?
    Solution
    For divisibility by 8, the last three digits should be divisible by 8.
    A can be 1, 3, 5, 7 or 9.
    There are 5 possible numbers, i.e. 7128, 7328, 7528, 7728 and 7928.
  • Question 11
    1 / -0
    What is the remainder when 927 is divided by 5?
    Solution
    Any odd power of 9 has 9 at the unit place.
    Thus, the remainder, when 927 is divided by 5, is 4.
  • Question 12
    1 / -0
    If 0.0ababab…. = , then find the value of ab.
    Solution
    x = 0.0ababab.….(1)
    10x = 0.ababab....(2)
    1000x = ab.abab….(3)
    Subtracting (2) from (3) ⇒ 990x = ab
    ⇒ x = =
    ∴ ab = 36
    Alternative method:
    0.0ababab… = =
    =
    ⇒ ab = 36
  • Question 13
    1 / -0
    A number, when divided by 296, leaves a remainder of 75. What will be the remainder obtained when the same number is divided by 37?
    Solution
    Let x be the number.
    Given: x = 296n + 75 [where n is the quotient]
    Or, x = 37 × 8n + 37 × 2 + 1
    Or, x = 37(8n + 2) + 1
    Hence, when x is divided by 37, it will give 1 as the remainder.
  • Question 14
    1 / -0
    Find the remainder when 32012 is divided by 13.
    Solution


    Thus, 9 is the remainder when 32012 is divided by 13.
  • Question 15
    1 / -0
    Find the remainder when 412 is divided by 57.
    Solution
    412 ÷ 57
    ⇒ (43)4 = (57 + 7)4
    Dividing by 57, we get remainder 74; but 74 can't be the remainder as it is more than 57.
    Now, apply the same theorem to 74.
    73 = 343
    73 ÷ 57, Remainder = 1
    Remainder when 73 × 7 is divided by 57,
    1 × 7 = 7
    So, the remainder is 7.
  • Question 16
    1 / -0
    What is the remainder when (1234567890123456789)24 is divided by 6561?
    Solution
    Sum of all the numerals of given number = 2 (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = 2 (45) = 90, i.e. this number is divisible by 9.
    Thus, we can write (1234567890123456789)24 as (9K)24. Similarly, 6561 can be written as 94.
    This means that the given number will be completely divisible by 6561, and remainder will be
    0.
  • Question 17
    1 / -0
    If x and y are two co-primes, then which of the following is true?
    Solution
    If x and y are co-primes, they don't have any common factor. Hence, their squares are also co- primes.
  • Question 18
    1 / -0
    Find the remainder when 721 + 722 + 723 + 724 is divided by 25.
    Solution
    721 + 722 + 723 + 724 is divided by 25, i.e.
    721(70 + 71 + 72 + 73)
    = 721(1 + 7 + 49 + 343)
    = 721(400)
    i.e. when 721(400) is divided by 25, remainder = 0 (as 400 is divisible by 25)
  • Question 19
    1 / -0
    There are two co-primes such that one number is exactly 8 times the other. What is the sum of the two numbers?
    Solution
    Co-primes are the numbers that do not have any common factor other than 1.
    So, the numbers must be 1 and 8.
    ∴ Sum = 1 + 8 = 9
  • Question 20
    1 / -0
    How many co-primes of 21 are less than 21?
    Solution
    Co-Prime of 21 less than 21 is 2,4,5,8,10,11,13,16,17 and 19.
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