Quant Boosters - Sibanand Pattnaik - Set 2



  • Direct formula for natural numbers = number of odd factors - 1
    So 3660 = 2^2 * 3 * 5 * 61
    so odd factors = 8
    Answer = 8 - 1 = 7
    as in case of "integers" its 2 * (number of odd factors) - 1
    had it been asked "integers then ur answer would have been 16 - 1 = 15

    Else you can think this way :

    3660 = 61 x 3 x4 x5
    3660= a + (a+1) + (a+2)...+ (a+(n-1)
    Then 3660 = n x Average of the above series
    If the number of terms in the series is odd, then the middle term of the series will be the average
    So say the number of terms is 61, then the middle term will be 3x4x5= 60. So the series will be: (30+31+32...60+61+62...90)
    If the number of terms is 3, the average will be 61x4x5= 1220. So the series is 1219+1220+1221.
    Therefore if the number of terms is odd, its easy to construct a series like this
    3660 has 7 odd factors other than 1



  • Q23) N has 6 factors and lies between 250 and 350. How many values of N are possible?
    a) 12
    b) 13
    c) 14
    d) None of the above



  • If a and b are prime numbers then N= a^5 or N= a^2 x b
    N= a^5 :No values
    N= a^2 x b
    If a= 2, then b can take prime values of 67, 71, 73, 79, 83: 5 values
    If a= 3, then b can take prime values of 29, 31, 37: 3 values
    If a= 5, then b can take prime values of 11, 13: 2 values
    Total values: 10



  • Q24) N^2 has 15 factors. M^2 has 9 factors. Which of the following cannot be the number of factors that MxN can have, if M and N are co prime to each other?
    a) 24
    b) 32
    c) 20
    d) 40



  • N= a^7 OR N= a^2xb^1
    M= p^4 OR M= p^1xq^1
    Where a,b, p and q are prime numbers
    Therefore M x N=
    i) a^7 x p^4 : 40 factors
    ii) a^2 x b x p^4 : 30 factors
    iii) a^7 x p x q : 32 factors
    iv) a^2 x b x p x q= 24 factors



  • Q25) Sum of HCF and LCM of 2 numbers is 30. How many such pairs of numbers exist?
    a) 6
    b) 4
    c) 7
    d) 8



  • h(1+xy) = 30
    h=1 xy=29 so 2^0=1
    h=2 xy= 14 = 2*7 so 2^1= 2
    h=3 xy=9 = 3^2 so 2^0= 1
    h=5 xy=5 so 2^0=1
    h=6 xy=4 so 2^0=1
    h=10 xy=2 so 2^0=1
    h=15 xy=1 so 1
    -> 1 + 2 + 5 * 1= 8



  • Q26) N is a natural number such that N/9 is a perfect square and N/4 is a perfect cube. Find how many factors does the smallest value of N have?
    a) 18
    b) 24
    c) 14
    d) 21



  • N = 2^2 * 3^6
    factors = 21



  • Q27) How many ways can 1500 be expressed as product of 3 integers ?



  • 1500 = 2^2 *3 * 5^3
    ordered solutions = 4c2 * 3c2 * 5c2 = 180
    integral : here we can have any 2 out 3 elements as negative ( for example , xyz = -x * -y * z)
    so 3c2 = 3 ways so total 180 * 3 = 540 negative solutions
    so finally 180 + 540 = 720 ordered integral solutions
    total perfect squares in 1500 = 4
    Total cubes = 1
    So (a,a,b) cases = 3 and (a,a,a) cases = 1
    now for integral (a,a,b), each of it can be (-a,-a,b) and (a,-a,-b)
    finally 2 * 4 -1 = 7 aab types and only 1 aaa type
    unordered = [720 - 3 * 7 -1]/6 + 8 = 124
    so its 124 ways



  • Q28) How many 2 digit numbers are there such that the sum of their digits is a prime number?
    a) 31
    b) 33
    c) 25
    d) 27



  • Sum of the digits is prime means the sum of the digits can be
    2: 11,20
    3:12,21,30
    5:41,14,23,32,50
    7:61,16,25,52,34,43,70
    11:92,29,83,38,74,47,65,56
    13:94,49,85,58,76,67
    17:98,89
    There are 33 such numbers. Hence (b)



  • Q29) N has 12 factors. Which of the following cannot be the number of factors N^2?
    a) 33
    b) 35
    c) 45
    d) 54



  • N can be of any of the following form:
    P^5x Q^1 therefore N^2 = P^10x Q^2; therefore N^2 has 33 factors
    P^3x Q^2 therefore N^2 = P^6x Q^4; therefore N^2 has 35 factors
    P^2x Q^1 x R^1 therefore N^2 = P^4x Q^2 x R^2; therefore N^2 has 45 factors
    Therefore 54 is the answer



  • Q30) How many factors (x, y, z) of 1000 exist such that HCF (x,y,z) = 1 ?



  • 1000 = 2^3 * 5^

    When we say HCF of factors (x,y,z) = 1 ; then we mostly talk of 3 cases ..

    Case -1 ; ( 1, 2^3, 5^3) ..
    now here ordered arrangements = 3^2*3! = 54

    Case - 2 ( 1, 1, 2^3) ; (1, 1, 5^3) ; (1, 1, 2^3 * 5^3)
    now total arrangement of these cases is 3!/2! ( 15) = 45

    case -3 when (1 , 1, 1) ; as we are taking of factors of 1000 with HCF 1 , so this has to be a case ... And it can be arranged in only n only 1 way .. so 1

    So finally 54 + 45 + 1 = 100


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