Number System Practice Gym by Anubhav Sehgal - Part 6


  • NMIMS, Mumbai (Marketing)


    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.

    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

    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

    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.

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

    N = 5a + 1 = 7b + 2 = 9c + 3 = 11d + 4
    2N = 5(2a) + 2 = 7(2b) + 4 = 9(2c) + 6 = 11(2d) + 8
    2N + 3 = 5(2a + 1) = 7(2b + 1) = 9(2c + 1) = 11(2d + 1)
    2N + 3 is the LCM of (5,7,9,11)
    N = [LCM(5,7,9,11) - 3]/2 = 1731

    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.

    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

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

    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

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

    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

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

    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.

    How many different marks are possible in CAT if there are 100 questions with marking scheme of +3 , -1 & 0 for correct, wrong & non attempt resp

    If no of marks are +n for correct and -1 for incorrect, then number of scores which are not possible is given by :
    (n - 1) + (n - 2) + .. + 1 = n(n - 1)/2
    Take +4 -1 scheme
    30 correct gives : 120
    29 correct gives : 116
    With 30 correct you can have 0 wrong
    So 1,2,3 wrong are not possible with 30 correct
    With 29 correct you can have 1 wrong
    But 2,3 Qs wrong is not..
    ..
    (n - 1) + (n - 2) + .. + 1 = n(n - 1)/2
    Which is nothing but C(n,2)

    How many positive integer solution for the equation x1 + x2 + x3 + … + xk =n, where 1 < = x1 < = x2 < = x3 … < = xk?

    a + b + c = 10
    1 < = a < = b < = c
    a = p + 1
    b = p + q + 1
    c = p + q + r + 1
    where p,q,r are non negative integers.
    3p + 2q + r = 7
    p = 0, 2q + r = 7 => 4 solutions
    p = 1, 2q + r = 4 => 3 solutions
    p = 2, 2q + r = 1 => 1 solution
    Total : 8 solutions

    Find the sum of all positive integral value a for which (2a + 124)/(a + 3) is an integer.

    2a + 124)/(a + 3)
    2(a + 62)/(a + 3)
    2[(a + 3 + 59)]/(a + 3)
    2 + 118/(a +3)
    a + 3 is a factor of 118
    118 = 2 * 59
    1,2,59,118
    a = -2,-1,56,115
    Sum of positive integral values = 171


Log in to reply
 

Looks like your connection to MBAtious was lost, please wait while we try to reconnect.