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

• 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?

• E(29) = 28
26^57 mod 29
26^56 * 26 mod 29
26

• Q11) How many scalene triangles are there for which the lengths of all sides are integers & the perimeter is 24 cm?

• For this particular case, [(n - 6)^2/48] where [.] nearest integer function.
[(24 - 6)^2 / 48 ] = 7

• Q12) The number of 4-letter words that can be formed out of the letters of the word FIITJEE is

• F,I-2,T,J,E-2
All different : 5c4 * 4! = 120
2 same 2 different : 2c1 * 4c2 * 4!/2! = 144
2 same 2 same : 2c2 * 4!/2!2! = 6
Total : 270

• Q13) Find the sum of the first 125 terms of sequence 1, 2, 1, 3, 2, 1, 4, 3, 2, 1, 5, 4, 3, 2 ...

• 1 ; 1 + 2 ; 1 + 2 + 3 ; ... 125 terms
120 terms if we have 1 ; 1,2 ; 1,2,3 ; .... ; 1,2,3,4,..,15
So,sum till 120 terms = n(n + 1)(2n + 1)/12 + n(n + 1)/4
where n = 15 => Sum = 620 + 60 = 680
After this,121st to 125th term : 16,15,14,13,12 = 70
Total : 750

• Q14) When a natural number is divided by 4 and 7, it gives remainder 3 and 2 respectively. What is the remainder obtained when same natural number is divided by 11?

• N = 4a + 3 = 7b + 2
Let us first find the general form of such a number using Chinese remainder theorem
4a + 3 = 7b + 2
4a = 7b – 1
a = 4b/4 + (3b – 1)/4
a = b + (3b – 1)/4
Pick the value of b for which (3b – 1)/4 is an integer.
b = 3 => a = 3 + 2 = 5
Number in general form = Value at any particular a, b + k * LCM (Coefficients of a, b) where k is any integer.
N = 23 + 28k
N mod 11 = (23 + 28k) mod 11 = (1 + 6k) mod 11 = can’t be determined uniquely as it would depend on value of k.

• Q15) Find the remainder when 1^1995 + 2^1995 + .... + 1996^1995 is divided by 1997

61

64

30

31

92

47

51

61