Number System Practice Gym by Anubhav Sehgal (NMIMS, Mumbai) - Part 11


  • NMIMS, Mumbai (Marketing)


    A natural number x leaves a remainder 1 when divided by p. The resultant quotient, when divided by q, leaves a remainder 2. The resultant quotient, when divided by r, leaves a remainder 3 and the quotient, thus obtained, is exactly divisible by 5. If p, q and r are all natural numbers, which of the following is the least possible value of x?

    It is a case of successive division. In such a case the number is written as:
    x= (p(q(r(5k) + 3) + 2) + 1.
    As the number needs to be smallest, we must take as small values as possible.
    Now, remainder with p is 1, so p >1. So, p = 2.
    Remainder with q = 2, so, q > 2, so q = 3.
    Remainder with r = 3, so, r > 3, so, r = 4.
    And, k = 0. x = 2(3(4(0) + 4) + 2) + 1 = 2(3(4) + 2) + 1 = 2(14) + 1 = 29.

    One of the smaller sides of a right angled triangle is (2^2) * (3^3) * (4^4) * (5^5) * (6^6) * (7^7). It is known that other two sides are integers. How many triangles of this type are possible?

    b^2 = (2^4) * (3^6) * (4^8 ) * (5^10) * (6^12) * (7^14)
    = (2^32) * (3^18 ) * (5^10) * (7^14)
    => (a + c)(a - c) = (2^32) * (3^18 ) * (5^10) * (7^14)
    Since both (a + c) and (a - c) are even, say 2k and 2n, then
    k * n = (2^30) * (3^18 ) * (5^10) * (7^14)

    Now, k is greater than n, so we just have to write (2^30) * (3^18 ) * (5^10) * (7^14) as product of two numbers where k is the greater one and n is smaller one, then we can get values for a and c
    So, answer will be (31 * 19 * 11 * 15 - 1)/2 = 48592

    In a test consisting of 15 questions, 3 marks are awarded for a correct answer, 1 mark is deducted for an incorrect answer and no mark is awarded for an unattempted question. If a student attempts at least one question in the paper, what is the number of distinct scores that he can get?

    Basically if no of marks are +n for correct and -1 for incorrect, then
    Number of score which are not possible is given by (n - 1) + (n - 2) + ... + 1 = n * (n - 1)/2 = C (n, 2)
    Assume a +6 and -1 marking scheme.
    When one gets 30 correct; he gets 180.
    And with 29 correct, he gets 174.
    So, for a score of more than174, one needs 30 correct.
    And with 30 correct i.e. all correct, you can have 0 wrong.
    So, 30 correct, 1 wrong or 2 wrong or 3 wrong or 4 wrong or 5 wrong is not possible.
    With 29 correct, 1 wrong is possible. But 2, 3 , 4 or 5 is not.
    With 28 correct, 3, 4, 5 wrong is not possible.
    So, basically, when we have +6 and -1: Number of invalid scores are: 1 + 2 + 3 + 4 + 5 = 15

    When we have +5 and -1:
    Number of invalid scores are: 1 + 2 + 3 + 4 = 10.

    When we have +4 and -1:
    Number of invalid scores are: 1+2+3 = 6.

    Here, 15 questions are there with +3 and -1 marking scheme.
    Highest marks possible = 15 * 3 = 45
    Lowest possible marks = 15 * -1 = -15
    Total marks = 45 – (-15) + 1 = 61 possible scores.
    Out of which, C (3, 2) = 3 are not possible.
    So, number of valid scores = 61 – C (3, 2) = 58.

    How many integers ‘n’ are there between 0 and 10^99, such that the unit digit of n^3 is 1?

    Each number from 1 to 10 results in distinct unit digit for cube.
    So, each digit in cube appears 1/10 times.
    So, from 1 to 10^99 it will appear 10^98 times

    How many subsets of the set {1, 2, 3, 4 ... 30} have the property that the sum of the elements of a subset is less than or equal to 232?

    Sum of all elements of set S = 465.
    For any subset having sum of elements x, there will be one complimentary set which is having sum of elements equal to (465 – x).

    Example let there be a set {1, 2, 3, 4}.
    Then subset which is having sum =3 will be {1, 2} {3}: (2)
    and subset having sum (1+2+3+4) -3 =7 will also be 2: {3,4} {1,2,4}

    So, number of subsets having sum of elements x = Number of subsets having sum of elements (465 - x)
    => No of subsets having sum 1 = No of subsets having sum 464
    Also, No of subsets having sum 2 = No of subsets having sum 463
    ….
    No of subsets having sum 232 = No of subsets having sum 233
    Adding them we will get,
    No of subsets having sum less than or equal to 232
    = No of subsets having sum more than 233
    = 2^30/2
    = 2^29

    When a number N (N>500) is successively divided by 3,5,7, it leaves remainder 2,3,6 respectively. Find the remainder when the smallest such N is divided by 35?

    N = 3(5(7k + 6) + 3) + 2 = 105k + 83.
    N = 83, 188, 293, 398, 503, etc.
    Our required N = 503.
    503 mod 35 = 13.

    If the first 99 natural numbers are written side by side to form a new number 12345678 …. 9899, then how many minimum numbers are to be removed so that, the new number is completely divisible by 11?

    Sum at odd places = 1+3+5+7+9+9(1+2+...+9) = 25+9(45) = 430
    Sum at even places = 2+4+6+8+10(1+2+.....+9) = 30+10(45) = 470
    Sum => even = 470 and odd => 430
    (Sum of digits at odd position from right) - (Sum of digits at even positions from right) = -40
    Even - Odd= 40
    Remove 99 -> diff =40
    Remove 98-> diff =41
    Remove 97-> diff =43
    Remove 96-> diff =46
    Remove 95-> diff =50
    Remove 94-> diff =55

    We should remove 6 numbers: {99, 98, 97, 96, 95, 94}.

    N = 4^7 × 5^9 + 2^4 × 7 + 3 × 5^3 + 2^6 × 5^8. How many distinct digits are there in the number N?

    = 32 * 10^9 + 84 + 375 + 25 * 10^6
    = 32025000459.
    So, 6 distinct digits.

    S1 = {2, 4, 6, 8, …, 800} S2 = {3, 6, 9, 12, ....., 900} If S3 = S1 ∪ S2, then what will be the 105th element of S3 if all its elements are arranged in increasing order?

    S1 ∪ S2 is a set comprising of all numbers which are multiples of 2 or 3 (or both).
    Now, we know that for a number N, there are N(1 - 1/2)(1 - 1/3) = N/3 numbers which are co-prime to 2 and 3 and less than N.

    So, there are 2N/3 numbers less than or equal to N which are multiples of 2 or 3 (or both).
    => 2N/3 = 106.
    => N = 159 is 106th such number.
    => 158 should be the 105th number.

    N is the smallest number such that N/2 is a perfect square and N/3 is a perfect cube. Find the number of factors of N.

    N/2 = x^2
    N/3 = y^3
    N^2 = 6.x^2.y^3
    N = xy.root(6y)
    For N to be a natural number,6y must be a perfect square => y = 6 => x = 2 * 3^2 and N = 2^3 * 3^4
    => Number of factors = 4 * 5 = 20


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