Quant Boosters  Sibanand Pattnaik  Set 2

Max power of "a" in the expression = 1+2+3+4+.....+20 = (20 * 21 )/2 = 210
To get coefficient of a^204 we have to find the constant term of expression having power a^6 or remove factors of a^6" , which are (x^11) * (x^55)" product of constant term = 5
 (x^22) * (x^44)" product of constant term = 8
 (x^11)" (x^22)" (x^33)" product of constant term = 6
 (x^6  6) * 1 .. product of constant terms = 6
so your required coefficient = 5 + 8  6  6 = 1

Q5) How many 5 digit numbers are there whose sum of digits is 12 ?

a + b + c + d + e = 12
now a should be atleast 1
(x+1) + b + c + d + e = 12
=> x + b + c + d + e = 11 so 15c4 casesnow remove the cases when x > = 9
so, (x + 9) + b + c + d + e = 11
=> x + b + c + d + e = 2 i.e 6c4 casesand when b, c , d , e > = 10 , we have to remove those cases as well .. isn't it ?
so , x + (b+10) + c + d + e = 11
=> x + b + c + d + e = 1 i.e 5c4 casesRequired numbers = 15c4  4 * 5c4  6c4
= 1365  20  15
= 1365  35
= 1330

Q6) There are exactly 2005 ordered pairs (x, y) of positive integers satisfying 1/x + 1/y = 1/N. Find the number of digits in smallest possible value of N if N is a natural number.

2005 = 5 * 401
N=a^p * b^q
N^2 = a^(2p) * b^(2q)
So (2p + 1)(2q + 1)= 5 * 401
p = 200, q = 2
N = 2^200 * 3^2
[LogN] + 1 = [200log2 + 2log3] + 1 = [61.16] + 1 = 62 digits

Q7) Total no of integral solution for x + y + z = 15 where 1≤ x, y, z ≤ 6

(6  a ) + (6  b) + (6  c) = 15
=> a + b + c = 3
5c2
Total no . of integral solutions = 10Alternate method : Coefficient of x^15 in (x + x^2 + x^3 + ... + x^6)^3
= x^12 in (1  x^6)^3 * (1x)^3
= 14c2  3c1 * 8c2 + 3c2
= 10

Q8) Find the coefficient of x^14 in (x^3 + x^6 + x^9 +...) * (x^2 + x^4 + x^6 + ...) * (x + x^2 + x^3 + ...)

Method  1 :
If you are good in binomial then do it directly. Else you can also think this in these lines. Is this not same as 3x + 2y + z = 14 and is this not same as finding the positive distinct integral solutions of the equation a + b + c = 14? so (13c2  3 * 6)/6 = 10Method 2 :
If you want to solve it by binomial then check below
coefficient of x^14 in (x^3 + x^6 + x^9 + ...) * (x^2 + x^4 + x^6 + ...) * (x + x^2 + x^3 + ...)
= coefficient of x^14  6 (1 + x^3 + x^6 + ...) * (1 + x^2 + x^4 + ...) * (1 + x + x^2 + x^3 + ...)
= coefficient of x^8 in (1 + x^2 + x^3 + x^4 + ... + x^8 +...) * (1 + x^2 + x^3 + x^4 + ... + x^8 + ....)
= 1 + 1 +1 + 1 + 1 + 2 + 1 + 2 = 10

Q9) In how many ways can an amount of rs. 100 be paid exactly using coins of denomination 1,2 and 5 such that atleast one coin of each denomination is used?

Basically the problem is to find +ve integer solutions to A + 2B + 5C = 100
So A + 2B can take values 5 , 15, 25 ,......95
The number of solutions corresponding to this will be 2 + 7 + 12 ... + 47 = 245
Also, A + 2B can be 10 , 20 , 30 ... 90
The number of solutions corresponding to this will be 4 + 9 + 14 ... + 44 = 216
so total = 245 + 216 = 461.

Q10) In how many ways can 2310 be expressed as a product of 3 factors?

2310 = 2 * 3 * 5 * 7 * 11
These 5 primes to be distributed into 3 places.
Total : 3^5 ways , but this is ordered ( arranged) and we need unordered.
We know that any triplet with distinct elements (a,b,c) is arranged in 3! ways, so all the distinct triplets in 3^5 = 243 are arranged in 3! Ways.
Except for one triplet (1, 1, 2310) which is arranged in 3!/2= 3 ways
So remove this from total and we are left with only distinct, is divide by 3! ,
So (3^5  3)/3! , and just add that one case of 1, 1, 2310
So (3^53)/3! + 1 = 41

Q11) How many pairs of positive integers (m, n) satisfy 1/m + 4/n = 1/12. Where n is an odd integer less than 60.

(m 12) * (n48) = 12 * 48 = 2^6 * 3^2
now for "n" to be odd , (n48) has to be odd
so n  48 can be either 1 or 3 or 3^2 (because we need ODD)
(m 12) * (n  48) = 2^6 * 3^2 * 1
OR (m 12) * (n  48) = 2^6 * 3 * 3
or (m 12) * (n  48) = 2^6 * 3^2
when n  48 = 1, n = 49 (acceptable)
when n  48 = 3, n = 51 (acceptable)
when n  48 = 3^2, n = 57 (acceptable)
only these 3 values are possible for n < 60 and odd

Q12) Find number of whole number solutions of a + b + c + d = 20 where a, b, c and d ≤ 8

Cofficient of x^20 in (1 + x + x^2 + x^3 + ... + x^8)^4
= coff of x^20 in (1  x^9)^4 * (1  x)^4
= 23c3  4 * 14c3 + 4c2 * 5c3
= 1771  1456 + 60
= 375For more detailed explanation of the concept, refer the video.

Q13) If k is a factor of (n+5) and (6n+54) where n is a natural number, how many possible values of k are there?
a) 6
b) 7
c) 8
d) 9

As given k must be a factor of 6n+30 and is also a factor of 6n+54.
So it must be a factor of (6n+54) – (6n+30) = 24 = 2^3 x 3
Therefore k can take 8 values
Hence (c)

Q14) The sum of all the factors of 480 including 1 and the number itself is 1512. What is the sum of the reciprocals of all the factors?
a) 3.15
b) 2.85
c) 2.55
d) 3.45