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+(n1) * 2) = 500
n^2 + (a1) * n = 500
here a is odd
so a1 will be even
and so (a1) * n will be even
so n^2 + even = even
so n^2 will be even so n will be even
now n = 500/n (a1)
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(1x^6/(1x) )^4
= x^4(1x^6)^4 (1x)^4
so coefficient of x^13 in x^4(1x^6)^4 (1x)^4
so coefficient of x^9 in (1x^6)^4 (1x)^4
(1x^6)^4 general term of this is 4cr * (1)^r * (x^6)^(4r)= 4cr * (1)^(4r) * x^(246r)
(1x)^4 coefficient of x^r1 is 4+r1c41=(3+r1)C3
so 4cr * (1)^(4r)*x^246r+r1 *( 3+r1)C3 so for coefficient of x^9
246r+r1=9 so 6rr1=15 so r =3 and r1=3 or r=4 and r1=9
so coefficient = 4cr * (1)^(4r) * ( 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 (x2) 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 (x2) 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 (x1)^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 (x1)Method 2 (common method for all kind of problem of this type)
x^2014 + 1 = P * (x1)^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 * (x1)^2 + mx + n
now put x=1
so m + n = 2
now differentiate it
2014 * x^(2013) = p(x) * 2(x1) + (x1)^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)