Quant Boosters  Anubhav Sehgal, NMIMS Mumbai  Set 3

We know that a digit or set of digits repeated (p – 1) times is divisible by p.
77777 … 36k times mod 37 = 0
Writing our number in extended form, (77777 … 70 times) * 10^2 + 77 mod 37 = 0
[This is nothing but 77777… 72 times ]
100r + 3 mod 37 = 0
26r mod 37 = 34
If you find difficulty in solving such equations for r, write it as the following and solve it like we do for a Chinese remainder theorem problem.
26r mod 37 = 34
=> 26r = 37k + 34
=> r = (26k + 26 + 11k + 8)/26
r = k + (11k + 8)/26
Find value of k such that (11k + 8)/26 is an integer.
Inside bracket values are 8, 19, 30, 41, [52]. Stop here as value is divisible by 26.
k = 4, => r = 4 + 52/26 = 6.

Q11) Find the remainder when 25^25^25 is divided by 9.

25 and 9 are coprime.
E(9) = 9 * (1 – 1/3) = 6
25^25 mod 6
= 1^25 mod 6 = 1.
So we have, 25^1 mod 9 = 7.

Q12) (18^2000 + 12^2000 – 5^2000  1) mod 221

(18^2000 – 5^2000) + (12^2000 – 1^2000) = B1 + B2
B1 and B2 both are divisible by 13 as (a^n – b^n) is divisible by (a – b) and (a + b) when n is even.
Also, (18^2000 – 1^2000) + (12^2000 – 5^2000) = B1 + B2
B1 and B2 both are divisible by 17 as (a^n – b^n) is divisible by (a – b) and (a + b) when n is even.
So the expression is divisible by 13 * 17 = 221.
Remainder will be zero.

Q13) 43^444 + 34^333 is divisible by?
a) 2
b) 5
c) 9
d) 11

Looking at the options, finding the unit digit will tell us divisibility for two of them (2 and 5).
Unit digit of 43^444 = Unit digit of 3^444 = Unit digit of 3^4 = 1
Unit digit of 34^333 = Unit digit of 4^333 = Unit digit of 4^1 = 4
Unit digit of expression = 1 + 4 = 5 => number is divisible by 5.The alternate way would have been to find remainder for each of the options.
a) 43^444 + 34^333 mod 2 = 1^444 + 0 mod 2 = 1; Not divisible.
b) 43^444 + 34^333 mod 9
E (9) = 6
(2)^444 + (2)^333 mod 9
444 mod 6 = 0
333 mod 6 = 3
so we get, 1 + (2)^3 mod 9 = 7 or 2.
Not divisible
c) 43^444 + 34^333 mod 11
(1)^444 + 1^333 mod 11
1 + 1 mod 11 = 2.
Not divisible.

Q14) Find the remainder when 1 × 2 + 2 × 3 + 3 × 4 + ... + 98 × 99 + 99 × 100 is divided by 101.

1 * 2 + 2 * 3 + 3 * 4 + … + 99 * 100 mod 101
2 + 6 + 12 +20 + … + 9900 mod 101
Let’s look at this series.
2  (4)  6  (6)  12  (8)  20  (10)  30 
The differences of the series are in AP. So this is called as a level 1 AP. General term for a level 1 AP is taken as a quadratic equation(cubic for level 2 and so on).
Tn = an^2 + bn + cT1 = a + b + c = 2
T2 = 4a + 2b + c = 6
T3 = 9a + 3b + c = 12T2 – T1 = 3a + b = 4
T3 – T2 = 5a + b = 6
2a = 2
a = 1
3a + b = 4 => b = 1
a + b + c = 2 => c = 0Hence, Tn = n^2 + n
Sn = n(n + 1)(2n + 1)/6 + n(n + 1)/2 = n(n + 1)/2 * [(2n + 1)/3 + 1] = n(n + 1)(n + 2)/3Our expression = S99 = 99 * 100 * 101/3 = 33 * 100 * 101 (Expression) mod 101 is therefore zero.

Q15) Find remainder when 1 * 1! + 2 * 2! + 3 * 3! + … + 10 * 10! is divided by 11.

1 * 1! + 2 * 2! + … + 10 * 10! Mod 11
(2 – 1)1! + (3 – 1)2! + …. + (11 – 1)10! Mod 11
2 * 1! – 1 * 1! + 3 * 2! – 1 * 2! + … + 11 * 10! – 1 * 10! Mod 11
11 * 10! – 1 * 1! Mod 11 [Rest all terms cancel out]
11! – 1 mod 11 = 1 or 10.

Q16) Find the remainder when 1^39 + 2^39 + 3^39 + 4^39 + ... + 12^39 is divided by 39.

We know that (a^n + b^n + … + k^n) is divisible by (a + b + c + .. + k) when n is odd and a, b, c, .. , k are in AP.
So, our expression is divisible by 12 * 13/2 = 78 and hence by extension by 39 since 39 is a factor of 78.
Remainder = 0.

Q17) Find remainder when 23^23 is divided by 53.

23^2 mod 53 = 1
23^23 mod 53 = (23^2)^11 * 23 mod 53 = 1^11 * 23 mod 53 = 23 or 30.

Q18) Sum of four two digit numbers is 256. None of the four numbers have a digit zero. Also all the eight digits are different. Which digit has not been used in the four numbers?

Here we use a new concept called the digital sum concept.
Digital sum is the recursive sum of digits until a single digit is obtained.
Example: 149 = 1 + 4 + 9 = 14 = 1 + 4 = 5. 5 is the digital sum of 149.Important usage of this concept
To check calculations. The digital sum must be same for your addition/subtraction/multiplication on both sides of your equation.Multiplying any number of 9 makes it digital sum of the result to be always 9.Example: 142 * 9 = 1278 = 1 + 2 + 7 + 8 = 18 = 1 + 8 = 9.Also, see here that, 1 + 4 + 2 = 7. 7 * 9 = 63 = 6 + 3 = 9 = Same as our result 1278’s digital sum.
Let us see how we will use this concept here to solve our problem.
We know that sum of four two digit numbers is 256 = 2 + 5 + 6 = 13 = 1 + 3 = 4 [digital sum].
Also, none of the numbers uses zero while all digits are distinct. Means we use 8 out of the 9 available digits as zero isn’t used at all.
Had we used all 9 digits somewhere, we would have had digital sum of LHS as (1 + 2 + .. + 9) = 4 * 9 = 9.
Due to the missing digit we are getting the digital sum as 4.
Hence the missing digit, x is given by 9 – x = 4 => x = 5.

Q19) What is the unit digit of [10^3000/(10^100 + 3)] where [.] is GIF.

Find remainder of (10^3000/(10^100 + 3))
Put x = 10^100 => x^30/(x + 3).
Put x = 3 => (3)^30 = 3^30 [Remainder Theorem]
Dividend = Quotient * Divisor + Remainder
Check for unit digit on both sides,
0 = x * 3 + 9 [ Dividend = 10^3000, Divisor = 10^100 + 3, Remainder = 3^30 , Quotient = unknown]
x = 7 [ Quotient’s unit digit is our answer]

Q20) Find the 50th digit after decimal in the expansion of 1/73.