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

• 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 = 0

Approach 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 5-digit-integer N so that 2N is also a 5-digit-integer and all digits 0, 1, 2, 3, ..., 9 are contained in both N and 2N

• Idea is that we have to maximize the digits.
4 has to be first digit of N. Giving us 9 as first digit of 2N ( 2 * 4 + 1 carryover)
Second digit of N cannot be 9, Since 9 is already first digit in 2N.
So let 8 be second digit of N. We have 2 * 8 = 16 +1 carryover = 17 giving 7 as the second digit of 2N.
Now let 6 be the 3rd digit of N. we have 2 * 6 = 12 + 1 carryover = 13 giving us 3 as the 3rd digit of 2N.
Now let 5 be the 4th digit of N. We have 2 * 5 = 10 giving us 0 as the 4th digit of 2N.
Finally we have 1 as the fifth digit of N giving us 2 as the last digit of 2N.
So the required numbers are : 48651 and 97302.

• Q26) Smallest whole number which when divided by 5,7,9 and 11 leaves the remainder 1,2,3 and 4 respectively.

34

65

61

107

68

43

145

89