Quant Boosters - Anubhav Sehgal, NMIMS Mumbai - Set 2


  • NMIMS, Mumbai (Marketing)


    24^1202 mod 1446
    1446 = 2 * 3 * 41
    Cyclicity of 1446 = LCM(1,2,240) = 240
    24^1202 mod 1446
    (24^240)^5 * 24^2 mod 1446
    1 * 576 mod 1446
    576


  • NMIMS, Mumbai (Marketing)


    Q22) Find the remainder when [102010!/(51005!^2) is divided by 101


  • NMIMS, Mumbai (Marketing)


    [102010!/(51005!^2) mod 101
    Direct application of Lucas Theorem
    Primarily used for finding remainders of large binomial coff with some prime p
    Here,
    [102010!/(51005!^2) = C(102010,51005)
    Since, 102010 = 10(101)^2 + 0(101) + 0
    or A00 in base 101 representation
    and
    51005 = 5(101)^2 + 0(101) + 0
    or 500 in base 101 representation
    C(102010,51005) mod 101
    = C(10,5) * C(0,0) * C(0,0)
    = 50 * 1 * 1 = 50


  • NMIMS, Mumbai (Marketing)


    Q23) Find (7^21 + 7^22 + 7^23 + 7^24) mod 25


  • NMIMS, Mumbai (Marketing)


    Approach 1 :
    (7^21 + 7^22 + 7^23 + 7^24) mod 25
    7^21(1 + 7 + 49 + 343) mod 25
    7^21 * 400 mod 25 = 0

    Approach 2 :
    E(25) = 20
    So,
    (7^21 + 7^22 + 7^23 + 7^24) mod 25
    (7 + 49 + 343 + 2401) mod 25
    7 - 1 - 7 + 1 = 0


  • NMIMS, Mumbai (Marketing)


    Q24) Let X be the largest positive number less than 10^6 such that when written in base 2, the binary representation consists of only 1s. Find the remainder when (X+2) is divided by 3.


  • NMIMS, Mumbai (Marketing)


    2^10 = 1024 > 10^3
    (2^10)^2 > 10^6
    => 2^19 is the largest power of 2 less than 10^6
    For a binary representation with all 1s, X must be
    2^19 - 1
    So (X + 2) mod 3 = (2^19 + 1) mod 3 = 0


  • NMIMS, Mumbai (Marketing)


    Q25) Find the LARGEST 5-digit-integer N so that 2N is also a 5-digit-integer and all digits 0, 1, 2, 3, ..., 9 are contained in both N and 2N


  • NMIMS, Mumbai (Marketing)


    Idea is that we have to maximize the digits.
    4 has to be first digit of N. Giving us 9 as first digit of 2N ( 2 * 4 + 1 carryover)
    Second digit of N cannot be 9, Since 9 is already first digit in 2N.
    So let 8 be second digit of N. We have 2 * 8 = 16 +1 carryover = 17 giving 7 as the second digit of 2N.
    Now let 6 be the 3rd digit of N. we have 2 * 6 = 12 + 1 carryover = 13 giving us 3 as the 3rd digit of 2N.
    Now let 5 be the 4th digit of N. We have 2 * 5 = 10 giving us 0 as the 4th digit of 2N.
    Finally we have 1 as the fifth digit of N giving us 2 as the last digit of 2N.
    So the required numbers are : 48651 and 97302.


  • NMIMS, Mumbai (Marketing)


    Q26) Smallest whole number which when divided by 5,7,9 and 11 leaves the remainder 1,2,3 and 4 respectively.


  • NMIMS, Mumbai (Marketing)


    Note here that 5 – 1 is not equal to 7 – 2 is not equal to 9 – 3 is not equal to 11 – 4.
    So you cannot apply LCM (5, 7, 9, 11) – N where N is the constant difference.
    N = 5a + 1
    N = 7b + 2
    N = 9c + 3
    N = 11d + 4

    Multiply each by 2 so that remainders also count by a gap of 2 like the divisors are doing.
    2N = 5(2a) + 2
    2N = 7(2b) + 4
    2N = 9(2c) + 6
    2N = 11(2d) + 8

    Now divisors and remainders share a common difference of 3. Designating them as new variables
    M = 5l + 2 = 7m + 4 = 9n + 6 = 11o + 8
    Least M = LCM (5, 7, 9, 11) – 3
    Least N = [LCM (5, 7, 9, 11) – 3]/2 = 1731 since M = 2N.

    Seeing the procedure you can now directly solve such questions in one line.
    Take LCM of divisors. Subtract the would be common remainder after you have figured out the number to multiply with to make the count of divisors and remainders of the same step.
    [Here Count of divisors was of 2 : 5, 7, 9, 11 and remainders was of 1. So we multiplied by 2/1 = 2]

    Another example : Divisors are 5,8,11 ; Remainders are 3,5,7. Multiplier = 3/2. See common remainder and then, Answer : [LCM(5,8,11) – 0.5]/(3/2) = 440 – 0.5/ (1.5) = 439.5/1.5 = 879/3 = 293

    NOTE : General form of number will be = Smallest number + k * LCM(Divisors). You may be asked to find the kth smallest number or some other variant. So you may add multiples of LCM to the smallest number to get the required answer.


  • NMIMS, Mumbai (Marketing)


    Q27) A positive integer p is called almost prime, when it has only 1 divisor aside from 1 and p. Find the sum of the 6 smallest almost primes.


  • NMIMS, Mumbai (Marketing)


    Only squares of primes have exactly 3 factors.
    Hence your required sum
    = 2^2 + 3^2 + 5^2 + 7^2 + 11^2 + 13^2 = 377


  • NMIMS, Mumbai (Marketing)


    Q28) Find n if sqrt(17^2 + 17^2 + ... n times) = 3 * 17^2


  • NMIMS, Mumbai (Marketing)


    3 * 17^2 = sqrt(9 * 17^4) = sqrt(9 * 17^2 * 17^2)
    i.e. 9 * 17^2 times 17^2 inside the square root.
    i.e 51^2 times
    i.e 2601


  • NMIMS, Mumbai (Marketing)


    Q29) a!b! = a! + b! and find (a + b). (Positive integers)


  • NMIMS, Mumbai (Marketing)


    General approach to solve such equations
    a!b! - a! - b! + 1 = 1
    (a! - 1)(b! - 1) = 1
    a! - 1 = b! - 1 = 1
    a! = b! = 2
    a = b = 2
    a + b = 4


  • NMIMS, Mumbai (Marketing)


    Q30) Which is the third smallest number when subtracted from 6300 results in a perfect square ?


  • NMIMS, Mumbai (Marketing)


    6300 - n = k^2
    80^2 = 6400
    79^2 will give us the first smallest number on subtracting..
    77^2 will give us the third
    6300 - 59|29 = 371.



  • @anubhav_sehgal

    79............? ??


 

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