Quant Boosters  Anubhav Sehgal, NMIMS Mumbai  Set 2

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?

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 4letter words that can be formed out of the letters of the word FIITJEE is

F,I2,T,J,E2
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

(a^n + b^n) is divisible by (a + b) when n is odd.
(1^1995 + 2^1995 + .... + 1996^1995) mod 1997
[(1^1995 + 1996^1995) + (2^1995 + 1995^1995) + ...] mod 1997
All brackets are divisible by 1997.
Hence remainder = 0

Q16) 10^x mod 13 = 1 and 1 < = x < = 100. How many values of x satisfy the given equation?