Quant Boosters - Anubhav Sehgal, NMIMS Mumbai - Set 3


  • NMIMS, Mumbai (Marketing)


    We know that (p – 2)! Mod p = 1 for any prime p.
    21! Mod 23 =1
    21 * 20 * 19 * 18 * 17! Mod 23 = 1
    (-2) * (-3) * (-4) * (-5) * 17! Mod 23 = 1
    [21 mod 23 = 21 or -2; In general, k mod n = k or (k – n) when k < n]
    120 * 17! Mod 23 = 1
    5r mod 23 = 1
    r = 14 as 70 mod 23 = 1.


  • NMIMS, Mumbai (Marketing)


    Q6) If f (x) = x^3 + x^2 + 2x – 1 is divided by (x + 2) gives remainder r and quotient q (x). Then, find the value of q (2) + r.


  • NMIMS, Mumbai (Marketing)


    Q7) Find remainder when 987987987… (123 digits) is divided by 1001.


  • NMIMS, Mumbai (Marketing)


    1001 = 10^3 + 1
    So, we may for triplets from right to left and take alternative signs + and –ve moving from right to left.
    123 digits means 41 triplets of three digits 987 each.
    When we take alternating sums from right to left, we will get,
    [(987 – 987) + (987 – 987) + … 20 such brackets] + 987 mod 1001 = 987.


  • NMIMS, Mumbai (Marketing)


    Q8) Find the remainder when 3^45 is divided by 125.


  • NMIMS, Mumbai (Marketing)


    The usual and the longer approach
    3^5 = 243. 243 mod 125 = 118 or -7
    3^45 mod 125 = (-7)^9 mod 125
    Now, 7^4 mod 125 = 2401 mod 125 = 26 [ You basically try to break it into powers which leave a smaller remainder with our divisor to make our calculations manageable. That’s why we didn’t take 7^2 =49, 7^3 = 343.]
    (-7)^4 * (-7)^4 * (-7) mod 125
    = 26 * 26 * -7 mod 125
    = 676 * -7 mod 125
    = 51 * -7 mod 125
    = -357 mod 125
    = 18 or -107.

    Smart approach [Will come from practice only]
    Your first instinct here could have been to use Euler. But when you notice E(125) = 100 and you are not getting near that power with multiples of 45, you decided to drop the idea. Some reverse engineering is being done here so note carefully the approach.
    3^45 mod 125 is to be found
    128 mod 125 = 3.
    So can we write 3^45 mod 125 = 128^45 mod 125? Sure we can but who thinks of going the other way round.
    What this does for us?
    128 = 2^7 thankfully and we get,
    3^45 mod 125
    = (2^7)^45 mod 125
    = 2^315 mod 125
    = 2^15 mod 125
    = 2^7 * 2^7 * 2 mod 125
    = 3 * 3 * 2 = 18.
    That simple by just going the other way round. A little out of the box thinking


  • NMIMS, Mumbai (Marketing)


    Remainder when f (x) is divided by (x – a) = f (a).
    r = f (-2) = -8 + 4 – 4 – 1 = -9
    q (x) = x^2 – x + 4 => q (2) = 4 – 2 + 4 = 6
    q (2) + r = -3


  • NMIMS, Mumbai (Marketing)


    Q9) Find the remainder when 2^2014 + 7 * 5^4020 is divided by 23


  • NMIMS, Mumbai (Marketing)


    E(23) = 22
    2014 mod 22 = 12
    4020 mod 22 = 16
    2^12 + 7 * 5^16 mod 23
    64 * 64 + 7 * 25^8 mod 23
    -5 * -5 + 7 * 2^8 mod 23
    25 + 7 * 256 mod 23
    = 25 + 21 mod 23
    = 0


  • NMIMS, Mumbai (Marketing)


    Q10) Find the remainder when 77777 … 70 times is divided by 37


  • NMIMS, Mumbai (Marketing)


    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.


  • NMIMS, Mumbai (Marketing)


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


  • NMIMS, Mumbai (Marketing)


    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.


  • NMIMS, Mumbai (Marketing)


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


  • NMIMS, Mumbai (Marketing)


    (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.


  • NMIMS, Mumbai (Marketing)


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


  • NMIMS, Mumbai (Marketing)


    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.


  • NMIMS, Mumbai (Marketing)


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


  • NMIMS, Mumbai (Marketing)


    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.


  • NMIMS, Mumbai (Marketing)


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


 

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