Number System Practice Gym by Anubhav Sehgal - Part 5


  • NMIMS, Mumbai (Marketing)


    Find the remainder when 1^1995 + 2^1995 + .... + 1996^1995 is divided by 1997

    (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

    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.

    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

    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

    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

    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

    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

    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

    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

    C(58,29) mod 29 = ?

    C(2n,n) = C(n,0)^2 + C(n,1)^2 + .. + C(n,n)^2
    For our question,
    C(58,29) mod 29
    = C(29,0)^2 + C(29,1)^2 + .. + C(29,29) mod 29
    All terms except first and last are divisible by 29.
    Remainder = C(29,0)^2 + C(29,29)^2
    = 1 + 1


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