# Quant Boosters - Sibanand Pattnaik - Set 2

• E = 1|x - 1| + 2|x - 1/2| + 3|x - 1/3| + ... + 20|x - 1/20|

All zeros occur at 1/n (n is a natural number less than 20)

Let say minimum of E occurs at x = 1/t

This is analogous to considering that there are 1 + 2 + 3 ... + 20 = 210 points on the number line. We need to select a point such that the sum of the distances of this point from all the 210 point will be a minimum. Thus, we need to take the median of these 210 points. i.e, any point between 105th point and 106th point. If we move to the left on the 105th point or to the right of the 106th point, the sum of the distances will increase.

Now the 105th point will be 1 + 2 + 3 + ... + 14 = 105

Thus the 105th point is 1/14 and the 106th point is 1/15. The expression, E will be minimum if we take x as any of these two values. i.e, at x = 1/14 or x = 1/15 (or any value in between)

• Q3) Solve for x if (x - 1/x)^1/2 + (1 - 1/x)^1/2 = x

• Domain x - 1/x > = 0,
1 - 1/x > = 0, x > 0
So (x - 1)/x > = 0 , (x^2 - 1)/x > 0
x > = 1 is Domain.
now squaring twice n simplifying
Replace x - 1/x = t
t^2 + 2 = x^2 + 1/x^2 so
t^2 + 2 - 2t - 1 = t^2 - 2t + 1 = (t-1)^2 = 0
So t = 1, x-1/x = 1
x^2 - 1 = x, x^2 - x - 1 = 0
x = ( 1 + root(5))/2

• Q4) what is the coefficient of a^204 in (a-1)(a^2 - 2) (a^3 -3)....(a^20 -20)

• Max power of "a" in the expression = 1+2+3+4+.....+20 = (20 * 21 )/2 = 210
To get co-efficient of a^204 we have to find the constant term of expression having power a^6 or remove factors of a^6" , which are

1. (x^1-1) * (x^5-5)" product of constant term = 5
2. (x^2-2) * (x^4-4)" product of constant term = 8
3. (x^1-1)" (x^2-2)" (x^3-3)" product of constant term = -6
4. (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 at-least 1
(x+1) + b + c + d + e = 12
=> x + b + c + d + e = 11 so 15c4 cases

now remove the cases when x > = 9
so, (x + 9) + b + c + d + e = 11
=> x + b + c + d + e = 2 i.e 6c4 cases

and 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 cases

Required 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 = 10

Alternate method : Coefficient of x^15 in (x + x^2 + x^3 + ... + x^6)^3
= x^12 in (1 - x^6)^3 * (1-x)^-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 = 10

Method 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 at-least 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 (3^5-3)/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) * (n-48) = 12 * 48 = 2^6 * 3^2
now for "n" to be odd , (n-48) 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

61

61

64

63

61

48

61

61