Remainder Theorem Practice Questions (Solved) - Hemant Malhotra


  • Director at ElitesGrid | CAT 2016 - QA : 99.94, LR-DI - 99.70% / XAT 2017 - QA : 99.975


    Find remainder when 6^83 + 8^83 divided by 49

    E(49)=49 * 6/7=42
    so 6^42 mod 49=1
    and 8^42 mod 49=1
    now 6^83=6^84 * (1/6)
    and 8^(83)=8^84/8
    so 1/6+1/8 mod 49
    14*(48^-1) mod 49
    now 48 mod 49=-1
    so 14 * (-1)^-1 mod 49
    -14 mod 49
    so 49 - 14=35

    OR if you want a much faster method,
    (7-1)^83+(7+1)^83
    expand this binomial expension
    even powers will get cancelled only odd will be there and 2 times
    so 2(7^83 + 83c1 * 7^81.........83c8227) mod 49
    now here all terms are div by 49 only 2 * 83 * 7 will not be div by 49
    so 2 * 83 * 7 mod 49=35

    3^33^333^3333^....^3333...(100 digits) mod 455 = ?

    3^33^333^3333.........100 digts mod 5
    E(5)=4
    so 33^333^333.^..100 digits mod 4=1
    so 3^1 mod 5=3
    now 3^33^333^333 .... mod 7
    E(7)=6
    33^333^3333^.... mod 6
    6 = 2 * 3
    33^333^333^333... mod 2=1
    33^333^3333... mod 3=0
    2a+1=3b
    so 3
    so 3^3 mod 7 =27 mod 7=6
    now for 13
    E(13)=12
    33^333^3333^... mod 12
    12=3*4
    33^333^3333.. mod 3=0
    and mod 4=1
    so 3a=4b+1 so 9 so 3^9 mod 13 so 27^3 mod 13=1
    so 5x+3=7y+6=13z+1 so 118

    Find Remainder when 2^100+ 2^200 + 2^300 +.....+ 2^100000 divided by 7

    2^100+4^100+8^100+16^100............2^1000)^100 mod 7
    now 2^100 mod 7=2
    4^100 mod 7=4
    and 8^100 mod 7=1
    now remainder in these three cases =2+4+1=7 mod 7=0
    so cycle will repeat like this in 3
    so total terms =
    2^1000=2*(2)^(n-1)
    2^1000=2^n
    n = 1000 terms
    so 1000 mod 3=1
    so only last term will count so remainder is 2

    if N=2222.....123 times then find Remainder when N is divided by 41

    Make group of 5 from right and add them if sum is div by 41 then num is div so 24 * 22222 + 222 mod 41 = 17
    (1 * 271 * 9 = 99999 = 10^5 - 1 that's y we make group of 5)

    Find remainder when 40! mod 83?

    82! mod 83=-1
    similarly 40!^2 * 41 * (-41)mod 83=-1
    40! * 41 mod 83=1
    if 40! mod 83=2
    then 2 * 41 mod 83=-1
    but -2 * 41mod 83=-82mod 83=1
    so it's -2=81
    rejected and -2 answer

    Find remainder when 19! is divided by number of digits in 23!

    21!, 22! ,23! and 24! have same digits so 19! mod 23 = 4

    19! mod 23=4
    21! mod 23=1
    so 21 * 20 * 19! mod 23 =1
    -3 * -2 * 19! mod 23=1
    6 * 19! mod 23=1
    6r = 23k + 1
    so r = 4

    or direct 6k-1=23 so k=4

    Find remainder when (105)^13! is divided by 103

    E(103)=102
    105^102 mod 103=1
    so now we have to check 13! will leave what value of remainder with 102
    13! mod 102
    102 = 6 * 17
    13! mod 6=0
    13! mod 17= 3
    6a = 17b + 3 so 54
    2^54 mod 103 = 8

    What is the remainder when ((55)^15!)^188 is divided by 17?

    15! is multiple of 16 and E(17)=,16 so 55^16k mod 17=1

    2nd approach - 55 mod 17=4
    (4^15!)^188 mod 17
    (16)^15!/2 )^188 mod 17
    (-1)^even mod 17 = 1

    What would be remainder when 8! is divided by 17

    2^7 * 3^2 * 5 * 7 mod 17
    (-8) * (-8) * 35 mod 17
    64 * 35 mod 17
    13

    Let 3 statements be made
    (P) 10^2p - 10^p + 1 is divisible by 13 for the largest integer p < 10
    (Q) The remainder on dividing 16! + 89 by 323 is q
    (R) 46C23 leaves remainder r on division by 23
    Then p + q + r equals
    (a) 12
    (b) 19
    (c) 26
    (d) 33
    (e) none of the foregoing

    a^2 - a + 1 mod 13 = 0
    a = 10^p
    a(max) = 10^9
    10^18 - 10^9 + 1 mod 13 = 1 + 1 + 1 = 3
    10^16 - 10^8 + 1 mod 13 = 3 - 9 + 1 = 8
    10^14 - 10^7 + 1 mod 13 = 9 + 3 + 1 = 13 mod13=0 so
    p = 7

    N = 16! + 89 mod 323
    323 = 17 * 19
    N mod 17 = -1 + 4 = 3
    N mod 19 = 9 + 13 = 3
    q = 3

    46c23 mod 23 = = 2=r

    p + q + r = 12


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