Quant Boosters  Anubhav Sehgal, NMIMS Mumbai  Set 2

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?

10^3 mod 13 = 1
10^6 mod 13 = 1
Hence 10^x mod 13 = 1 for x = 6,12,18,...,96
16 values.

Q17) Find 123234345456567678789 mod 37