Quant Boosters  Sibanand Pattnaik  Set 2

N= a^7 OR N= a^2xb^1
M= p^4 OR M= p^1xq^1
Where a,b, p and q are prime numbers
Therefore M x N=
i) a^7 x p^4 : 40 factors
ii) a^2 x b x p^4 : 30 factors
iii) a^7 x p x q : 32 factors
iv) a^2 x b x p x q= 24 factors

Q25) Sum of HCF and LCM of 2 numbers is 30. How many such pairs of numbers exist?
a) 6
b) 4
c) 7
d) 8

h(1+xy) = 30
h=1 xy=29 so 2^0=1
h=2 xy= 14 = 2*7 so 2^1= 2
h=3 xy=9 = 3^2 so 2^0= 1
h=5 xy=5 so 2^0=1
h=6 xy=4 so 2^0=1
h=10 xy=2 so 2^0=1
h=15 xy=1 so 1
> 1 + 2 + 5 * 1= 8

Q26) N is a natural number such that N/9 is a perfect square and N/4 is a perfect cube. Find how many factors does the smallest value of N have?
a) 18
b) 24
c) 14
d) 21

N = 2^2 * 3^6
factors = 21

Q27) How many ways can 1500 be expressed as product of 3 integers ?

1500 = 2^2 *3 * 5^3
ordered solutions = 4c2 * 3c2 * 5c2 = 180
integral : here we can have any 2 out 3 elements as negative ( for example , xyz = x * y * z)
so 3c2 = 3 ways so total 180 * 3 = 540 negative solutions
so finally 180 + 540 = 720 ordered integral solutions
total perfect squares in 1500 = 4
Total cubes = 1
So (a,a,b) cases = 3 and (a,a,a) cases = 1
now for integral (a,a,b), each of it can be (a,a,b) and (a,a,b)
finally 2 * 4 1 = 7 aab types and only 1 aaa type
unordered = [720  3 * 7 1]/6 + 8 = 124
so its 124 ways

Q28) How many 2 digit numbers are there such that the sum of their digits is a prime number?
a) 31
b) 33
c) 25
d) 27

Sum of the digits is prime means the sum of the digits can be
2: 11,20
3:12,21,30
5:41,14,23,32,50
7:61,16,25,52,34,43,70
11:92,29,83,38,74,47,65,56
13:94,49,85,58,76,67
17:98,89
There are 33 such numbers. Hence (b)

Q29) N has 12 factors. Which of the following cannot be the number of factors N^2?
a) 33
b) 35
c) 45
d) 54

N can be of any of the following form:
P^5x Q^1 therefore N^2 = P^10x Q^2; therefore N^2 has 33 factors
P^3x Q^2 therefore N^2 = P^6x Q^4; therefore N^2 has 35 factors
P^2x Q^1 x R^1 therefore N^2 = P^4x Q^2 x R^2; therefore N^2 has 45 factors
Therefore 54 is the answer

Q30) How many factors (x, y, z) of 1000 exist such that HCF (x,y,z) = 1 ?

1000 = 2^3 * 5^
When we say HCF of factors (x,y,z) = 1 ; then we mostly talk of 3 cases ..
Case 1 ; ( 1, 2^3, 5^3) ..
now here ordered arrangements = 3^2*3! = 54Case  2 ( 1, 1, 2^3) ; (1, 1, 5^3) ; (1, 1, 2^3 * 5^3)
now total arrangement of these cases is 3!/2! ( 15) = 45case 3 when (1 , 1, 1) ; as we are taking of factors of 1000 with HCF 1 , so this has to be a case ... And it can be arranged in only n only 1 way .. so 1
So finally 54 + 45 + 1 = 100