# Quant Boosters - Hemant Malhotra - Set 11

• 648=2^3 * 3^4
so x+y+z+k=3 so 6c3=20
and x+y+z+k=4 =7c3=35
so total positive =700
now cases can be
all positive=700
2 positive 2 negative =4c2 * 700
or all negative =700
so 700+700+6 * 700
4200+1400=5600
a^x * b^y * c^z * d^k = 2^3 * 3^4
now a ,b,c,d all will contain 0,1,2,or 3 power of 2 and sum of all those will be equal to 3
so x+y+z+k=3
and same for 3
so x+y+z+k=4

• Q24) In how many ways can 5^17 be written as a product of three positive numbers

• a + b + c = 17
19C2 = 171
when two are same :
2a + c = 17
total 9 solutions
(171 - 3 * 9)/ 3! + 9 = 33

• Q25) |x * y * z| = 210. Find number of Integral Solutions

• 210= 2 * 3 * 5 * 7
So positive = 3 * 3 * 3 * 3 = 3^4
Now each of three can be either positive or negative
So 2 * 2 * 2 * 3^4 = 8 * 3^4
Unordered: 81 - 3 / 6 + 1 = 14

• Q26) In how many ways 1001^2 can be written as product of 3 integers (Ordered and unordered)

• a * b * c = 7^2 * 11^2 * 13^2
x+y+z=2=4c2=6
x+y+z=2=4c2=6
x+y+z=2=4c2=6
so 6^3+3 * 6^3
4 * 6^3=864
now unordered
2 same
a^2 * b=1^2 * x
7^2 * x
11^2 * x
13^2 * x
(7 * 11)^2 * x
(7 * 13)^2 * x
(11 * 13)^2 * x
(7 * 11 * 13)^2
same 8 for negative so 16
so 864-16 * 3/6 +16

• Q27) In how many ways 1000 can b expressed as the product of 3 integers? (ordered and unordered)

• 1000=2^3 * 5^3
so x+y+z=3
so 5c2=10
and x+y+z=3
so 10
so 10 * 10=100
and negative 3 * 100=300
so total =400

now unordered
same a^2 * b=2^3 * 5^3
1^2 * x
2^2 * x
5^2 * x
10^2 * x
(-1)^2 * x
(-2)^2 * x
(-5)^2 * x
(-10)^2 * x
so total y=8
but one case will be all same here so y=7
and z=1
so 6x + 3 * 7 + 1=400
so x=400-22/6
so unordered=x+y+z

• Q28) Find coefficient of x^3 * y^4 * z^3 in expansion of (2x + y + 4z)^10

• general term of this equation is = 10!/(r1! * r2! * r3!) * (2x)^r1 * (y)^r2 * (4z)^r3
= 10!/(r1! * r2! * r3!) * (2)^r1 * (4)^r3 * x^r1 * y^r2 * z^r3
now we have to find x^3 * y^4 * z^3 so r1=3 ,r2=4 and r3=3
put tvalue of r1, r2, r3 here 10!/(r1! * r2! * r3!) * (2)^r1 * (4)^r3 * x^r1 * y^r2 * z^r3
so coefficient is 10!/(3! * 4! * 3!) * (2)^3 * (4)^3

• Q29) Find the largest coefficient in (x + y + z + k)^15

• general term = 15!/(r1! * r2! * r3! * r4!) * x^r1 * y^r2 * z^r3 * k^r4
so basically we have to maximize 15!/(r1! * r2! * r3! * r4!) term
so we have to minimize (r1! * r2! * r3! * r4!) so for this symmetry is useful so put values as close as possible here
r1 = r2 = r3 = 4 and r4 = 3
so largest coefficient is 15!/(3! * (4!)^3)

• Q30) Find the coefficient of x^6 in (1 + x^2 - x^3)^8

• general term = 8!/r1! * r2! * r3! * (1)^r1 * (x^2)^r2 * (-x^3)^r3
= 8!/r1! * r2! * r3! * x^(2r1) * (-1)^r3 * (x)^3r3
= 8!/r1! * r2! * r3! * (-1)^r3 * x^(2r2+3r3)
now we have to find coefficient of x^6 here (for any coefficient same procedure )
so r1 + r2 + r3 = 8
2r2 + 3r3 = 6
find r1, r2 and r3 here
let r3=0 so r2=3 so r1=2 and let r3=2,so r2=0 so r1=6 (r1,r2,r3 should be integer because term can't be negative)
now coefficient will be 8!/r1! * r2! * r3! *(-1)^r3
put both values of r1,r2 and r3 and add
so 8!/5! * 3!+8!/2! * 6!=84

• @hemant_malhotra as the no. of real roots is asked so it will always be 3

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