Quant Boosters  Anubhav Sehgal, NMIMS Mumbai  Set 2

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

123234345456567678789 mod 37
10^3 mod 37 = 1
Hence we can form groups of 3 digits from right,
Find individual remainders, and
Add them up for final remainder.
123 mod 37 = 12
234 mod 37 = 12
..
789 mod 37 = 12
7 * 12 mod 37 = 10 = Answer

Q18) Find 10111213141516171819 mod 33

10111213141516171819 mod 33
10^2 mod 33 = 1
Hence we can form groups of 2 digits from right,
Find individual remainders, and
Add them up for final remainder
(10 + 11 + 12 + .. + 19) mod 33
145 mod 33 = 13

Q19) Find (13^3 + 14^3 + ... + 34^3) mod 35

(13^3 + 14^3 + ... + 34^3) mod 35
[34 * 35/2]^2  [12 * 13/2]^2 mod 35
[35*17]^2  [78]^2 mod 35
(0  64) mod 35
29 or 6

Q20) When a natural number M is divided by 4 and 7, it gives remainder 3 and 2 respectively. Even when another natural number N is divided by 4 and 7, it also gives remainder 3 and 2 respectively. There is no other number in between M and N exhibiting these properties. P is a natural number that gives the same remainder after dividing each of M and N. How many values of P are possible?

4a + 3 = 7b + 2
a = 5, b = 3
Number = 23 + 28k
Let M = 23 + 28a
and N = 23 + 28b
Since P leaves same remainder with each M and N
Therefore P must be a factor of 28 since constant(23) will leave same remainder.
Final remainder will depend on 28k term which can be same for both M and N
if it is a factor of 28.
1,2,4,7,14,28 : 6 Values

Q21) Find 24^1202 mod 1446

24^1202 mod 1446
1446 = 2 * 3 * 41
Cyclicity of 1446 = LCM(1,2,240) = 240
24^1202 mod 1446
(24^240)^5 * 24^2 mod 1446
1 * 576 mod 1446
576

Q22) Find the remainder when [102010!/(51005!^2) is divided by 101