Quant Boosters - Hemant Malhotra - Set 9



  • HCF(6N+1,15N+2))
    HCF(12N + 2, 15N + 2)
    HCF(12N + 2, 3N)
    HCF(12N + 2, 2)
    HCF(6n + 1, 1)
    so all numbers will give hcf as 1 so 99



  • Q22) How many Ways 199 can be represented as sum of 3 distinct positive integers ?



  • a + b + c = 199
    we need positive integers so
    a = a'+1
    b = b'+1
    c = c'+1
    so a' + b' + c' = 196
    now total cases 196+3-1c3-1=198c2
    now this will include cases when all r different let those cases are x
    and way to arrange them is 6 so 6x
    now case when 2 are same and one is different
    2a' + b' = 196
    b' = 196 - 2a'
    so 99 cases
    and way to arrange them is 3!/2!=3
    so 99 * 3 =
    now when all are same
    so 3a' = 196
    a' = 196/3 whihc is not possible so no case
    now total ordered = 6x + 3 * 99
    so x = 198c2 - 3 * 99/6



  • Q23) A=333....333 (51 digits) and B= 666....666 (51 digits). Find 52nd digit (counting from right) in the product A * B
    a) 9
    b) 7
    c) 2
    d) 1
    e) none of the above



  • Check out the pattern,
    3 * 6 = 18 so 2nd right digit is 1
    33 * 66 = 2178 so 3rd right is 1
    same in 51 digits, 52nd will be 1



  • Q24) What will be remainder when (67^67 + 67) is divided by 68



  • 67 mod 68 = -1
    so (-1)^67 - 1
    so -2
    so 68 - 2 = 66 remainder



  • Q25) I went to 'murugan idly shop' to get packed breakfast for my guests.There were 4 types of idli available and i wanted to take atleast 1 pack of each and atmost 5 packs of any.If I bought a total of 10 packs of idly,in how many ways could I have been billed?
    a) 80
    b) 84
    c) 68
    d) 56



  • a+b+c+d=10
    1+a'+1+b'+1+c'+1+d'=10
    so a'+b'+c'+d'=6
    here 0 < =a' < =4
    6+4-1c4-1=9c3
    now a'=5+a''
    5+a''+b'+c'+d'=6
    so a''+b;+c'+d'=1
    so 4c3
    same for all 4 so 4 * 4c3=16
    so 9c3-16
    9 * 8 * 7/3 * 2 - 16
    84 - 16 = 68



  • Q26) If a and b are natural numbers with no common prime factor and c is the greatest common divisor of (a+ b) and (a^2 +b^2) then how many values can c take?
    a) 0
    b) 1
    c) 2
    d) 3
    e) Cannot be determined



  • gcd(a,b) = 1.
    a^2+b^2= (a+b)^2-2ab
    now c will divide either 2 or ab
    Case 1:c divided ab
    so c either divides a or b ,If c divides a it must divide a as well
    so c will divide a and b both so c=1
    Case 2: c divides 2.
    c=1 or 2
    so value of c could be 1 or 2



  • Q27) Digital sum of 1! + 2! + ... + 10 !



  • 1!+2!+3! mod 9=0
    4!+5!+6!
    4!*(1+5+6)
    4!*12 mod9=0
    7!+8!=7!(1+8 ) so mod9=0
    9!+10! multiple of 9=0
    so digital sum=9



  • Q28) a + b + c + d = 21, number of solutions such that a, b, c and d are distinct natural numbers.



  • Method 1 :
    a=x+1
    b=x+y+2
    c=x+y+z+3
    d=x+y+z+w+4
    so 4x + 3y + 2z + w = 11
    (6+5+3+2) + (4+3+1) + (2+1) = 27
    So, 24*27 = 648 solutions
    This is just simple counting

    Method 2 :
    a+b+c+d=21
    so a'+1+b'+1+c'+1+d'+1=21
    so a'+b'+c'+d'=17
    so total solution =17+4-1c4-1=20c3
    now we want all distinct cases
    so remove cases
    case1- when three values are equal
    so a=b=c
    so 3a+d=20
    so a wiill vary from 1 to 6 so 6 soulution
    and total ways to arrange them 4!/3!=4
    case2- when two are equal
    so a=b
    2a+c+d=20
    so 45 cases but this will include cases when three are equal so remove that cases so 39 cases a
    and way to arrange them 4!/2!*2!=12
    so 20c3 - 4 * 6 - 12 * 39 = 648



  • Q29) Consider a pair (a,b) of natural numbers satisfying a + b^2 +c^3 = abc where c is the greatest common divisor of a and b .Then , how many such pairs are possible?
    (a) 2
    (b) 3
    (c) 4
    (d) 5
    (e) 6



  • let a = Mc, b = Nc, so that M and N are coprime.
    Then: Mc + N^2c^2 +c^3 = MNc^3.
    M + N^2c + c^2 = MNc^2
    So c must divide M. Put M = M'c, then M' + N^2 + c = M'Nc^2.
    So M' = (N^2 + c)/(Nc^2 - 1).
    So M'c^2 = N + (N+c^3)/(Nc^2 - 1).
    So (N + c^3)/(Nc^2 - 1) is an integer.
    If c = 1, (N+1)/(N-1) can only be an integer for N = 2 or 3. so (a, b) = (5, 2) and (5, 3).

    let c > 1
    bcz (N + c^3)/(Nc^2 - 1) is an integer and positive, we must have N + c^3 > = Ng^2 - 1, so N < = (c^3 + 1)/(c^2 - 1). If c = 2, then N < = 3. Then N = 1 gives the solution (a, b) = (4, 2), N = 2 gives (N + c^3)/(Nc^2 - 1) non-integral and hence no solution,

    N = 3 gives the solution (a, b) = (4, 6). and for c > 2 there will be no solution so number of values = 4



  • Q30) The positive integer N has exactly six distinct integer divisors inclusing 1 and N. The product of five of these is 648. Which of the following must be a divisor of N
    a. 4
    b. 9
    c. 12
    d. 16
    e. 24



  • Method 1 -
    Product of factors = N (number of factors/2))
    let 6th factor = k
    so k * 648 = N^3
    so k * 3^4 * 2^3 = N^3
    RHS is prefect cube so LHS shuld be perfect cube
    so k = 3^2 = 9

    Method2 : N = 2 * 3^2
    648 = 1 * 2 * 3 * 6 * 18
    remaining 9 so OA = 9



  • @hemant_malhotra I don't understand how 248 came


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