Quant Boosters  Anubhav Sehgal, NMIMS Mumbai  Set 2

(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

[102010!/(51005!^2) mod 101
Direct application of Lucas Theorem
Primarily used for finding remainders of large binomial coff with some prime p
Here,
[102010!/(51005!^2) = C(102010,51005)
Since, 102010 = 10(101)^2 + 0(101) + 0
or A00 in base 101 representation
and
51005 = 5(101)^2 + 0(101) + 0
or 500 in base 101 representation
C(102010,51005) mod 101
= C(10,5) * C(0,0) * C(0,0)
= 50 * 1 * 1 = 50

Q23) Find (7^21 + 7^22 + 7^23 + 7^24) mod 25

Approach 1 :
(7^21 + 7^22 + 7^23 + 7^24) mod 25
7^21(1 + 7 + 49 + 343) mod 25
7^21 * 400 mod 25 = 0Approach 2 :
E(25) = 20
So,
(7^21 + 7^22 + 7^23 + 7^24) mod 25
(7 + 49 + 343 + 2401) mod 25
7  1  7 + 1 = 0

Q24) Let X be the largest positive number less than 10^6 such that when written in base 2, the binary representation consists of only 1s. Find the remainder when (X+2) is divided by 3.

2^10 = 1024 > 10^3
(2^10)^2 > 10^6
=> 2^19 is the largest power of 2 less than 10^6
For a binary representation with all 1s, X must be
2^19  1
So (X + 2) mod 3 = (2^19 + 1) mod 3 = 0

Q25) Find the LARGEST 5digitinteger N so that 2N is also a 5digitinteger and all digits 0, 1, 2, 3, ..., 9 are contained in both N and 2N