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.

Answer : 19
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 = 161,209,221Case 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 = 125Sum = 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 1100 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?