Quant Boosters - Sibanand Pattnaik - Set 2


  • QA/DILR Mentor | Be Legend


    Highest power of 5 in 125!= 31.
    Highest power of 5 in 140!=34
    But highest power of 5 in 141!, 142!, 143! And 144! Is also 34.
    So N can be 140 to 144, any of these values
    However the highest power of 7 in any of these values is same i.e. 22


  • QA/DILR Mentor | Be Legend


    Q17) A! + B! = N^2, Where A, B and N are whole numbers. How many possible combinations are there?
    a) 1
    b) 2
    c) 3
    d) 4


  • QA/DILR Mentor | Be Legend


    0! + 4! = 5^2
    1! + 4! = 5^2
    0! + 5! = 11^2
    1! + 5! = 11^2
    Hence 4 combinations are possible


  • QA/DILR Mentor | Be Legend


    Q18) x, y and z are natural numbers such that x > y > z > 1 and PRODUCT= 3003. Let A and B be the maximum and minimum values of x + y + z. Find A-B.
    a) 152
    b) 45
    c) 107
    d) None of the above


  • QA/DILR Mentor | Be Legend


    3003= 3 x 7 x 11 x 13
    A= 143+7+3= 153
    B= 21 + 11 + 13= 45
    A-B= 108
    Hence (d)


  • QA/DILR Mentor | Be Legend


    Q19) How many natural numbers less than 4444 are "not co primes" to 247?
    a) 585
    b) 341
    c) 244
    d) 557


  • QA/DILR Mentor | Be Legend


    247= 13 x 19
    So all multiples of 13 and 19 has to be excluded till 4444
    Multiples of 13 : 341
    Multiples of 19: 233
    Multiple of both 13 and 19 i.e. of 247: 17
    Therefore number of numbers less than 4444 which are co-prime to 247: 341+233-17= 557


  • QA/DILR Mentor | Be Legend


    Q20) N , a natural number has 2 prime factors. N^3 can have which of the following number of factors?
    a) 18
    b) 90
    c) 70
    d) 84


  • QA/DILR Mentor | Be Legend


    n^3 means exponent should be a multiple of 3
    70= 10 * 7 i.e (9+1) * (6+1) both powers have to be multiple of 3


  • QA/DILR Mentor | Be Legend


    Q21) Let D be the number of divisors of a natural number N. Let D1 be the number of divisors of D, D2 the number of divisors of D1 and so on. What will be the final value of Dn?
    a) 1
    b) 2
    c) 3
    d) None of the above


  • QA/DILR Mentor | Be Legend


    You will finally end up with a prime number which will have 2 divisors. 2 has 2 divisors. Hence (b)


  • QA/DILR Mentor | Be Legend


    Q22) In how many ways can 3660 be written as the sum of 2 or more consecutive positive integers?
    a) 2
    b) 3
    c) 4
    d) More than 4


  • QA/DILR Mentor | Be Legend


    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


  • QA/DILR Mentor | Be Legend


    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


  • QA/DILR Mentor | Be Legend


    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


  • QA/DILR Mentor | Be Legend


    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


  • QA/DILR Mentor | Be Legend


    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


  • QA/DILR Mentor | Be Legend


    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


  • QA/DILR Mentor | Be Legend


    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


  • QA/DILR Mentor | Be Legend


    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


 

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