Quant with RR - Number Theory



  • What is the least multiple of 7, which when divided by 2, 3, 4, 5 and 6 leaves the remainders 1, 2, 3, 4 and 5 respectively?

    N/2 = 1 or -1
    N/3 = 2 or -1
    N/4 = 3 or -1
    N/5 = 4 or -1
    N/6 = 5 or -1
    you can see that -1 is constant
    Formula to find number
    N = LCM of (2,3,4,5,6) k + constant
    N = 60k -1

    Now
    60k-1 / 7 should be divisible
    To find K lets simplify it more and make remainder 0
    4k-1/7
    Now obviously k should be 2 to make it perfectly divisible
    4 (2)-1/7
    = 0 rem

    We Found K = 2
    Put it in that number
    N = 60k - 1
    N = 60 (2) - 1
    N = 119

    What is the remainder when (2^100+3^100+4^100 5^100) is divided by 7?

    2^100 + 3^100 + 4^100 + 5^100/7
    = (2^3)^33 * 2 + (3^3)^33 * 3 + (4^3)^33 * 3 + (5^3)^33 * 3 / 7
    = 2 * 8^33 + 3 * 27^33 + 4 * 64^33 + 5 * 125^33 / 7
    = 2 * 1^33 + 3 * -1^33 + 4 * 1^33 + 5 * -1^33 / 7
    = 2 * 1 + 3 * -1 + 4 * 1 + 5 * -1 / 7
    = 2 - 3 + 4 -5 / 7
    = -2/7
    = 7-2
    = 5 remainder

    If a number is divided by 14 and the remainder is 5, then if the same number is divided by 7, what is the remainder?

    N/14 = 5 remainder
    N could be 19, 33, 47,....
    Least number is 19
    So 19/7 = 5 remainder

    Or

    N = 14k + 5
    Now
    N/7
    14k+5/7
    0*k+5/7
    5/7
    5 remainder

    Find the remainder when 17^53 is divided by 27

    Euler of 27 = 18
    So Euler says
    17^18/27 = 1 remainder
    17^36/27 = 1
    .
    Similarly 17^54/27 = 1
    17^53 * 17 /27 = 1
    R * 17 = 27k+1
    R = 27k+1/17
    Let K be any number which perfectly divides the equation
    R = 27 (5)+1/17
    R = 8 remainder

    Find the remainder when 5^100000 is divided by 196

    5^100000/196
    = 5^100000/ 4*49

    Lets divide it separately
    5^100000/4
    = 1^100000/4
    = 1 remainder

    5^100000/49
    Euler of 49 = 49 (1 - 1/7) = 49 (6/7) = 42
    Euler says
    5^42/49 = 1 remainder
    5^84/49 = also 1

    Same way
    5^100002/49 = 1
    5^100000 * 5^2/49 = 1
    R*25 = 49k+1
    R = 49k+1/25
    Let k be any lesser value which perfectly divides the equation
    Let k be 1
    R = 50/25 = 2 remainder

    So our answer is that number which leaves remainder 1 when divided by 4 & remainder 2 when divided by 49
    Apply Chinese remainder theorem and get your answer

    Answer is 149

    Find the remainder when 2017^22 is divided by 23

    Fermat thereom says that
    When a and p are coprime to each other
    a^(p-1)/p = 1 remainder
    Where, p is a prime number
    Here 23 is prime
    So
    2017^22/23
    = 1 remainder

    What is the unit digit of 123^4567! * 765^4321?

    123^4567! * 765^4321
    We have to find unit digit so we are concerned of unit digit only
    3^4567! * 5^4321
    (3^4)^(4567!/4) * 5^4321
    (xxxx1)^(4567!/4) * 5^4321
    [ Note :- 1^n = 1 unit and 5^n = 5 unit]
    1^(4567!/4) * 5^4321
    = 1 * 5
    = 5 unit digit

    What is the remainder when (49^15 - 1) is divided by 14?

    49^15 - 1/14
    = (7^2)^15 - 1/14
    = 7^30 - 1/14
    [ Note :- 7^n/14 = 7 remainder ]
    = 7-1/14
    = 6/14
    = 6 remainder

    What is the remainder when 2000^2001 is divided by 26?

    2000^2001/26
    24^2001/26
    26 = 13*2

    Divide separately
    24^2001/13
    -2^2001/13
    -2 * 2^2000/13
    -2 * (2^6)^333 * 2^2/13
    -8 * 64^333/13
    -8 * -1^333/13
    (-1 ^odd = - 1 )
    -8*-1/13
    8/13
    8 remainder

    24^2001/2
    = 0 remainder

    So our answer is that number which leaves remainder 8 when divided by 12 & remainder 0 when divided by 2

    N/13 = 8
    N could be 8, 21, 29,...

    N/ 2 = 0
    N could be 0,2,4,6,8,...

    The very first number common in both term is 8

    So final remainder is 8

    Answer 8

    Or
    24^2001/26
    -2^2001/26
    Eulee of 26 is 12
    (-2^12)^166 * -2^9/26
    1 * -2^9/26
    -512/26
    -18/26
    26-18
    8 remainder

    When 3n is divided by 7, the remainder is 4. what is the remainder when 2n is divided by 7?

    3n/7 = 4
    3N could be 4, 11, 18..
    First multiple of 3 here is 18
    3n = 18
    N = 6

    Now
    2n/7
    = 2 (6)/7
    = 12/7
    = 5 remainder

    Find the remainder when 1^5 + 2^5 + 3^5 +.......+ 100^5 is divided by 4

    Every even no. here is perfectly divisible by 4
    Our question shorts to
    1^5 +3^5 +5^5 +.....+99^5/4

    Euler of 4 = 2^2
    4 (1 - 1/2) = 4 (1/2) = 2
    so euler says
    a^2/4 = 0 rem
    a^4/4 = 0 rem
    .
    Where "a" and 4 are coprime to each other

    Our question again shorts to
    1+3+5+....99/4
    Sum of odd natural number/4
    = n^2/4
    [ Where N is the term, Here N is 50th term ]
    = 50^2/4
    = 2500/4
    = 0 remainder

    What is the remainder of 10! + 1111111…99times when divided by 1331?

    10! + 11111....(99 times)/1331

    Find remainder separately

    10!/1331
    = 494 remainder ----> (1)

    1111.... (99 times)/1331

    I found this pattern for every odd N > 3

    Remainder of [10(n-1/2)^2 + (n-1/2)(n-3/2)]/121 * 11 + 1

    So if question is
    11111/1331 here N is 5
    R of [10 (5-1/2)^2+(2)(1)]/121 *11 +1
    = R of [10 (2^2)+2]/121 *11+1
    = [42] 11+1 = 463

    If 1111111/1331
    N is 7
    = R of [10 (3^2)+(3)(2)]/121 * 11+1
    = [96]11 +1 = 1057

    Similarly in this question
    N is 99
    = R of [10 (49^2)+(49)(48)]/121 * 11+1
    = R of [10 (2401)+2352]/121 *11+1
    = R of [10 (102)+53]/121 *11+1
    = R of [1073/121] * 11+1
    = [105] 11+1
    = 1156 remainder --->(2)

    So now merge it again
    10! + 111... (99 times)/1331
    = 494 + 1156/1331
    = 1650/1331
    = 319 ----> final remainder

    Answer is 319

    What will be the remainder when(1234567890123456789)^24 is divided by 6561?

    Remainder of Ka/Kb
    = K * rem of (a/b)

    (1234567890123456789)^24/6561
    = (9K)^24 / 9^4
    = (9^24 * K^24) / 9^4
    = 9^4 * rem of (9^20 * K^24 / 0)
    = 9^4 * 0
    = 0 remainder

    How many numbers between 100 and 200 (including both 100 and 200), are not divisible by any of these numbers: 2, 3 and 5?

    Euler of 2 * 3 * 5
    = (1- 1/2)(1 - 1/3)(1 - 1/5)
    = (1/2)(2/3)(4/5)
    = 4/15

    Number not divisible :-

    Between 1-100
    100 (4/15)
    = 26

    Between 1-195
    = 195 (4/15)
    = 52

    So from 1-195
    52-26 = 26

    Between 196-200 only 2 number is not divisible I.e 197 & 199

    Total number
    = 26+2
    = 28 < --- answer

    What is the greatest 4 digit no when divided by 10,11,15 & 22 leaves 3,4,8 & 15 as remainders respectively?

    N/10 = 3 or -7
    N/11 = 4 or -7
    N/15 = 8 or -7
    N/22 = 15 or -7
    Format is
    Lcm of ( 10,11,15,22 ) k + constant
    = 330k-7
    If we put k as 1 then we will get least no.
    =330 (1)-7
    = 323 is the least no. Which satisfy above condition
    But we want highest 4 digit no.
    So hit and trial method
    Let k be 30
    = 330 (30)-7
    = 9893
    But if we take k as 31 it will give 5 digit no. But we want highest 4 digit no.
    So ans is 9893

    What is the remainder if 13571357… (upto 1000 digits) is divided by 101?

    1357/101 = 44 remainder
    13571357/101 = 44 * 2 = 88 remainder
    135713571357/101 = 44 * 3 = 132/101 = 31 remainder
    so make block of 4 number I.e (1357)
    It will give 44 as remainder
    Here 1000/4 = 250 such blocks
    So 44 * 250/101
    = 44 * 48/101
    = 92 remainder

    What is the remainder when 987987987… (123 digits) is divided by 1001?

    987987987.... (123digit)/1001
    Remember
    987987/1001 = 0 remainder
    Because
    987987/1001 can be written as
    987*1001/1001 = 0 remainder
    So 987987 I.e make a block of 6 digit and that will perfectly divided by 1001
    Here 123/6 = 20 such block but remain 3 digit I.e 987
    So our question shorts to
    987/1001
    = 987 remainder

    If the LCM of two natural numbers is 300, how many sets of A and B are possible?

    Lcm (a, b) = 300 = 2^2 * 3* 5^2

    Let us think from the perspective of each prime.

    First of all, both ‘a’ and ‘b’ must be factors of 2^2
    So we have 3 possibilities of both ‘a’ and ‘b’. Hence 3^2 possibilities.
    But from these possibilities we must remove those cases where ‘a’ and ‘b’ are neither 2^2. This will happen when both ‘a’ and ‘b’ are factors of 2^1.
    So that 2^2 cases must be removed.
    So we have (3^2 - 2^2) cases where lcm will be 2^2

    Similarly for 3^1 case
    We will have (2^2 - 1^2)

    Similarly for 5^2 case
    We will have (3^2 - 2^2)

    Total ordered cases
    = (3^2 - 2^2)(2^2 - 1^2)(3^2 - 2^2)
    = 5 * 3 * 5
    = 75 cases

    In order to convert to unordered solution, we must first understand that except for the case (300,300) all cases have been counted twice.

    So, the number of unordered solutions will be (75+1)/2
    = 38 cases

    In a list of 200 numbers, each number after the first is 4 more than the number that comes before it. What is the difference between the first and the last number on the list?

    T1 = a
    T2 = a+4
    T3 = a+8
    T4 = a+12
    .
    .
    If you observe clearly you will see difference from 1st term follows:-
    Difference = (Tn -1) 4
    = (200-1) 4
    = 796

    What is the remainder when 47^37^27 is divided by 11?

    47^37^27/11
    Euler of prime P is (P-1)
    Euler of 11 is 10
    Power/euler
    37^27/10
    = 7^27/10
    = (7^2)^13 * 7/10
    = 49^13 * 7/10
    = (-1)^13 * 7/10
    [ Note :- -1^odd = -1 ]
    = -1 * 7/10
    = -7/10
    = 10-7
    = 3 remainder

    Our question shorts to
    47^3/11
    = 3^3/11
    = 27/11
    = 5 ---> final remainder

    Last two digit of 21^50 - 8?

    21^50 - 8
    Last two digit of (a1)^xxb
    = last digit of (a * b) 1
    = last digit of (2 * 0) 1
    = xxxx01
    = 01 ---> last two digit of 21^50
    Now
    Last two digit of 21^50 - 8
    = xxx01 - 8
    = 93


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