# Quant Boosters - Hemant Malhotra - Set 7

• x^n-a^n=(x-a)(x^(n-1)+x^(n-2) + ... a^(n-1)
f(x^12)=1+(x^12)+(x^12)^2+(x^12)^3+(x^12)^4+(x^12)^5
f(x^12)=1+(((x^12 -1)+1+(x^12)^2 -1 +1 +((x^12)^3 -1 +1 +(x^12)^4 -1 +1 +(x^12)^5 -1+1
f(x^12)=6+((x^12-1)+(x^12)^2-1+(x^12)^3 -1+(x^12)^4 -1 +(x^12)^5-1
now x^12-1=(x^6-1)(x^6+1)
x^6-1=(x-1)(1+x+x^2+x^3+x^4+x^5)
so every term will contain x^6-1 and will be divisible by 1+x+x^2+x^3+x^4+x^5
so OA=6

• Q20) Two different positive numbers x and y such that each differ from their reciprocals by 1. Find x + y

• x and y will be roots of |a-1/a|=1
so a-1/a=1 or a-1/a=-1
a^2-a-1=0 or a^2+a-1=0
a=(1+-√(5))/2 or a=(-1+-√(5))/2
so 4 values and sum of x and y will be (1+√5)/2 + (-1+√5)/2= √5

• Q21) Let f be a one-to one function from the set of natural numbers to itself such that f(mn) = f(m) x f(n) for all natural numbers m and n. What is the least possible value of f(111)

• one to one so single image for every element
now f(1 * 1)=f(1) * f(1)
so f(1)(f(1)-1)=0
so f(1)=1 bcz f(1)=0 not possible we are taking about natural number
now f(111)=f(3 * 37) = f(3) * f(37)
we want f(3) and f(37) as min as possible and different from each other
so put f(3)=2 and f(37)=3
so 2 * 3=6

• Q22) If a three digit number ‘abc’ has 3 factors, how many factors does the 6-digit number ‘abcabc’ have?

• square of prime
so abc = k^2
now abcabc = abc * 1000 + abc
abc * 1001
so k^2 * 7 * 11 * 13
now when k=any prime other that 11,13 till 31 then 24 factors
when k is prime 11 or 13 let k=11 then 11^3 * 7 * 13 = 4 * 2 * 2 = 16

• 3I + 2F = I + F
so F = -2I
so 0 < = F < 1
so 0 < = -2I < 1
so -1/2 < I < = 0
so I = 0 is possibility
so F = 0
so x = 0 is only value

• Q24) A fare dice is thrown 5 times. What is the probability that a prime number appear even number of times

• 2,3,5 three prime numbers
so out of 6 we have 3 prime number
so prob of prime=1/2 and prob of non prime=1/2
so 5c0 * (1/2)^0 * (1/2)^55c2 * (1/2)^2 * (1/2)^3 + 5c4 * (1/2)^4 * (1/2)
132 + 10 * 1/4 * 1/8 + 5 * 1/16 * 1/2
16/32=1/2

• Q25) Find the largest integer N , if N + 496 = a^2 , N + 224 = b^2.

• a^2-496=b^2-224
so (a-b)(a+b)=272
now N will be max when a and b are as max possible
a=69 and b=67
so N=69^2-496=4265

• Q26) 2x + y + z = 20. Find number of positive integral solution.

• y and z both even then y=2k and z=2m
then x+k+m=10 so 9c2=36 cases
when y and z both odd then y=2m-1 and z=2k-1
then 2x+2m+2k=22
so x+m+k=11 so 10c2=45
so 36+45=81

• Q27) In a right angled triangle with integral sides, smallest side is four times of difference of other two sides and sum of all sides is 120 then find area of this triangle
a. 60
b. 120
c. 180
d. 240
e. None of the above

• Method 1
S = 4 * (H-M)
Smallest side is multiple of 4 and S+M+H=120 so triplet will be
S+M+(S+4M)/4=120
so 5S+8M=480
so M=(60)-((5S/8))
so S=24 and M=45 so H=51 so area=540

Method 2
S = 4 * (H-M)
so 8 = 4 * (17-15)
so multiple of this will give our ans
so 24, 45, 51 will be triplet

• Q28) gcd(1,55) + gcd(2,55) + ... + gcd(55,55) = ?

• E(55)=40
now factors of 5 and 11
till 50 , multiple of 5 are 10 so 5 * 10 = 50
and multiple of 11 will be 11,22,33,44 so 11 * 4 = 44
now 55 also
so 40 + 50 + 44 + 55
= 90 + 99 = 189

• Q29) For how many integral x, (3x+8)^2/(3x-2) is also an integer

54

61

61

71

61

61

61