Quant Boosters - Hemant Malhotra - Set 1



  • Q27) How many sets of three or more consecutive odd numbers can be formed such that their sum is 500?
    a) 10
    b) 0
    c) 3
    d) 4
    e) 5



  • odd numbers with d = 2
    sum = 500
    n/2 * (2a+(n-1) * 2) = 500
    n^2 + (a-1) * n = 500
    here a is odd
    so a-1 will be even
    and so (a-1) * n will be even
    so n^2 + even = even
    so n^2 will be even so n will be even
    now n = 500/n- (a-1)
    we want n =even
    500 = 2^2 * 5^3
    so find even factors
    (2,250) and (10,50) (50,10) and (250,2)
    here first term is n
    n is greater than 2 so 1st case rejected so 3 case possible



  • Q28) Find coefficient of x^13 in (x+x^2+x^3+x^4+x^5+x^6)^4



  • (x + x^2 + x^3 + x^4 + x^5 + x^6)^4
    = x^4(1 + x + x^2 + x^3 + x^4 + x^5)^4
    =x^4(1-x^6/(1-x) )^4
    = x^4(1-x^6)^4 (1-x)^-4
    so coefficient of x^13 in x^4(1-x^6)^4 (1-x)^-4
    so coefficient of x^9 in (1-x^6)^4 (1-x)^-4
    (1-x^6)^4 general term of this is 4cr * (1)^r * (-x^6)^(4-r)= 4cr * (-1)^(4-r) * x^(24-6r)
    (1-x)^-4 coefficient of x^r1 is 4+r-1c4-1=(3+r1)C3
    so 4cr * (-1)^(4-r)*x^24-6r+r1 *( 3+r1)C3 so for coefficient of x^9
    24-6r+r1=9 so 6r-r1=15 so r =3 and r1=3 or r=4 and r1=9
    so coefficient = 4cr * (-1)^(4-r) * ( 3+r1)C3 ..
    = 4c3 * (-1)^1 * 6c3 + 4c4 * (-1)^0 * (3+9)c3
    = 140



  • Q29) Let P(x) be a polynomial in x. When P(x) is divided by (x-2) the remainder is 8. When P(x) is divided by (x+2) the remainder is 4. What is the remainder when P(x) is divided by x^2 - 4



  • When P(x) is divided by (x-2) the remainder is 8 so remainder will be of form ax+b
    so P(x)=Q(x) * (x^2 - 4 ) + (ax+b)
    so P(x)=Q(x) (x - 2)(x + 2) + (ax+b)
    so put x = 2 in remainder that will be equal to 8
    so 2a + b = 8
    now put x = -2 then remainder will be 4
    and -2a + b = 4
    solve these two equations
    so b=6 and a=1
    so remainder = x + 6



  • Q30) Find the remainder when x^2014 + 1 is divided by x^2 - 2x + 1



  • Method - 1
    x^2014 mod (x-1)^2
    let x - 1 = a
    so (1+a)^2014 +1 mod a^2
    = 1 + (1 + 2004 * a + 2004 * a^2 + ... 2004 * a^2014) mod a^2
    so 1 + (1 + 2004 * a)
    2 + 2004 * a
    2 + 2004 (x-1)

    Method 2- (common method for all kind of problem of this type)
    x^2014 + 1 = P * (x-1)^2 + mx + n
    where mx + n is remainder because divisor power is x^2 so remainder will be less than that so
    now
    x^2014 + 1 = P * (x-1)^2 + mx + n
    now put x=1
    so m + n = 2
    now differentiate it
    2014 * x^(2013) = p(x) * 2(x-1) + (x-1)^2 * p'(x) + m
    now again put x=1
    2014 = m
    so m + n = 2
    so n = -2012
    so remainder mx+n
    2014x - 2012



  • @hemant_malhotra One doubt.. In this 243 we are also including that case when the subset will be null. because from all 5 pairs "none" can also be selected..
    So answer 243 - 1 = 242. Please clear my doubt sir.



  • @pratik0809 It is asked to create subset so that NO two elements sum to 11. So Null set satisfies right ?

    Logic here is, for a usual case (without any conditions), we can have 2^n subsets for a set of n elements. Why ? because each element can either be present or not in the subset. So each element can have two options giving a total of 2^n possibilities.

    Here, we have 5 pairs and each pair can contribute in 3 ways - First element selected or Second element selected or None is selected.

    So total subsets = 3^5 = 243 ways (which includes null set too)


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