# Quant Boosters - Anubhav Sehgal, NMIMS Mumbai - Set 2

• Number of Questions - 30
Topic - Number Theory
Solved ? - Yes
Source -

• Q1) Find the number of composite factors of 420.

Concept : Total number of factors = 1 + (Prime factors) + (Composite Factors)

• Q2) For how many positive integers x between 1 and 1000, both inclusive, is 4x^6 + x^3 + 5 divisible by 7?

• x^3 mod 7 = 0,1,-1 only
Case 1 : (4x^6 + x^3 + 5) mod 7 = 5
Case 2 : (4x^6 + x^3 + 5) mod 7 = 4(1) + 1 + 5 mod 7 = 10 mod 7 = 3
Case 3 : (4x^6 + x^3 + 5) mod 7 = 4(1) - 1 + 5 mod 7 = 8 mod 7 = 1
Hence zero values.

• Q3) The value of (222) in base ‘X’ when converted to base 10 is ‘P’. The value of (222) in base ‘Y’ when converted to base 10 is Q. If (P – Q) in base 10 = 28, then what is the value of (Q – X ) in base 10?

• Solve and Share your method for this one

• Q4) X is a set of all natural numbers with 4 factors such that sum of all its factors excluding number itself is 31. Find the sum of all such numbers

• Case 1 : N = a * b
Factors will be 1,a,b,ab
1 + a + b = 31
a + b = 30
a,b = 7,23 ; 11,19 ; 13,17 => N = 161,209,221

Case 2 : N = a^3
Factors : 1,a,a^2,a^3
1 + a + a^2 = 31
a(a + 1) = 30
a = 5 is the only case
N = 125

Sum = 716

• Q5) Find the remainder when C(58,29) is divided by 29.

• C(2n,n) = C(n,0)^2 + C(n,1)^2 + .. + C(n,n)^2
For our question,
C(58,29) mod 29
= C(29,0)^2 + C(29,1)^2 + .. + C(29,29) mod 29
All terms except first and last are divisible by 29.
Remainder = C(29,0)^2 + C(29,29)^2
= 1 + 1

• Q6) How many numbers from 1 to 100 are NOT divisible by 2 or 3 or 5 or 7 ?

• 105 * 1/2 * 2/3 * 4/5 * 6/7 - {101,103} = 22

Why 105 ?

Step 1 : Find numbers not divisible by 2,3,5 or 7 by simple Euler number's application.
Step 2 : Remove numbers like 103 which being not divisible by 2,3,5 or 7 were removed in euler but do not belong to our range of 1-100 for which answer is to be calculated.
105 Chosen as it is the closest(to100) multiple of LCM of uncancelled denominators 5,7

• Q7) HCF of 2 numbers is 12 and their sum if 144. Find the maximum value of their product

• 12(a + b) = 144
a + b = 12
As near to each other as possible with they being coprime to each other.
5, 7
Max product = (12 * 5) * (12 * 7) = 60 * 84 = 5040.

• Q8) Highest power of 8 in 17! + 18! + 19! + . . . + 100!?

• 17!(1 + 18 + 18 * 19 + ... 18 * 19 * 20 * ..100)
17!(1 + even)
17! * odd
Find power of 2 and subsequently 8 in 17!
Power of 2 : 8 + 4 + 2 + 1 = 15
Power of 8 : [15/3] = 5

• Q9) What is the highest power of 3 available in the expression 58! - 38!

• 38!(58 * 57 * .. * 39 - 1)
Power of 3 in 38! = 12 + 4 + 1 = 17
Power of 3 inside bracket = 0
Highest power to divide 58! - 38! = 17

• Q10) Remainder when 26^57 is divided by 29?

61

65

61

106

61

61

58

61