# 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 co-prime.
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 + c

T1 = a + b + c = 2
T2 = 4a + 2b + c = 6
T3 = 9a + 3b + c = 12

T2 – T1 = 3a + b = 4
T3 – T2 = 5a + b = 6
2a = 2
a = 1
3a + b = 4 => b = 1
a + b + c = 2 => c = 0

Hence, 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)/3

Our 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.

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