Base System - Vikas Saini



  • We use base 10 for mathematical operation. So we use 10 digit (0-9). After 9, we use for next number 10. Then 11,12,13.....19. Then again 20,21....and so on.

    So we use digits as per that number system.

    In base 2 - 2 digits (0,1)

    base 3 -  3 digits (0,1,2)

    base 4 -  4 digits (0,1,2,3)

    base 10 -  10 digits (0,1,2,3,4,5,6,7,8,9)

    Base 13 - 13 digits (0,1,2,3,4,5,6,7,8,9,A,B,C,D)

    base 16 - 16 digits(0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F)

    Cream of the piece -

    1. In any base system n, number of digits is equal to n.

    2. The max digit in any number system 'n' is 'n-1' .

    Conversion in any base 'n' from base '10'

    Convert 77(base 10) in base 6.

    77 = (abc)_6

    77 = a x 6^2 + b x 6 + c.

    if a = 1,

    then b = 5, c = 5.

    (155)_6 = 71.(not equal to 77).

    if a=2, b = 0, c = 5.

    (205)_6 = (77)_10.

    Convert (158 )_10 in base 16.

    (158 )_10 = (abc)_16.

    a x 16^2 + b x 16 + c.

    here a = 0, b = 9, c = E.

    (9E)_16 = (158 )_10.

    Convert (1078 )_10 in base 7.

    (1078 )_10 = (abcd)_7

    a x 7^3 + b x 7^2 + c x 7 + d.

    a = 3, b = 1, c = 0, d = 0.

    (3100)_7 = 1078.

    Conversion from base 'n' to base '10'

    (abcd)_n = a x n^3 + b x n^2 + c x n^3 + d.

    Convert (66)_7 in base 10.

    6 x 7 + 6 = 48.

    (66)_7  = (48 )_10.

    Convert (599)_11 in base 10.

    5 x 11^2 + 9 x 11 + 9

    = 605 + 99 +9

    = 713.

    Convert (1274)_9 in base 10.

    1 x 9^3 + 2 x 9^2 + 7 x 9 + 4

    = 729 + 162 + 63 + 4

    = 958.

    (1274)_9  = (958 )_10.

    Now some questions from  Base system.

    Q. 677 has exactly 5 digits when converted into base 'n' from the decimal system. What is the minimum possible value of n.

    Solution :-

    Let's start from base 3.

    smallest 5 digit  number in base 3 = (10000)_3 = 3^4.

    Largest 5 digit number in base 3 = (22222)_3 = 2(11111)_3 = 3^5 - 1.

    in base 3 not possible.

    Now let's take base 4.

    smallest number = (10000)_4 = 4^4 = 256.

    Largest number = (33333)_4 = 4^5 - 1 = 1023.

    256 < 677 < 1023.

    Hence base 4.

    Q. How many 3 digit numbers are there in base 9 system.

    Solution :-

    Hundred place = 1 to 8 = 8 digit.

    Ten's place = 0 to 8 = 9 digit.

    Unit's place = 0 to 8 = 9 digit.

    Total = 8 x 9 x 9 = 648 digits.

    Q. How many decimal numbers are three digit numbers in base 7 and four digit numbers in base 6.

    Solution :-

    Smallest 3 digit number in base 7 = (100)_7 = 7^2 = (49)_10.

    Largest 3 digit number in base 7 = (666)_7 = 7^3 - 1 = (342)_10.

     [49,342] numbers are 3 digits numbers in base 7.

    Smallest 4 digit number in base 6 = 6^3 = 216.

    Largest 4 digit number in base 6 = 6^4 - 1 = 1295.

    [216,1295] numbers are 4 digit numbers in base 6.

    [216,342] common numbers.

    Total decimal numbers = 342 - 216 + 1 = 127.

    Q. (N)_10 = (aaa)_3.
    here a is a single digit number, then N is a multiple of.

    (a) 9                     (b)3a
    (c)13                    (d)26.

    Solution :-

    (aaa)_3 = a (111)_3

    a (3^2 + 3 + 1)

    (13a)_10 = (N)_10.

    Hence 13.

    Q. How many 3 digits are there in decimal system which have exactly 3 digits when expressed in both base 7 & base 13.

    Solve:-

    smallest 3 digit number in base 7 = (100)_7 = 7^2 = 49.

    Largest 3 digit number in base 7 = (666)_7 = 7^3 - 1 = 342.

    [100,342] are the number which are exactly 3 digit number in base 7.

    smallest 3 digit number in base 13 = (100)_13 = 13^2 = 169.

    Largest 3 digit numbers in base 13 = (CCC)_13 = 12^3 - 1 = 1727.

    [169,999] are the three digit numbers which are exactly 3 digit numbers in base 13.

    [ 169,342] is common in both.

    Total = 342 - 169 + 1 = 174 numbers.

    Q. (206)_x + (216)_x = (424)_x .

    what is value of x ?

    Solution :-

    2x^2 + 6 + 2x^2 + x + 6 = 4x^2 + 2x +4

    = > x + 12 = 2x + 4.

    = > x = 8.

    Q. (xy)_8 + (yx)_8 = (abc)_10.

    How many values are possible of abc.

    Solution :-

    8x + y + 8y + x = abc.

    9x + 9y = abc.

    9(x+y) = abc.

    x = 5, y = 7 then abc = 108.

    x = 6, y =7 then abc = 117.

    but abc are distinct digits.

    so only 1 value 108.

    Q. A two digit number in base 7 is equal to the two digit number formed by reversing it digits but in base 13.

    How many such numbers are possible ?

    Solution :-

    (xy)_7  = (yx)_13

    7x+y = 13y+x

    6x = 12y

    x = 2y.

    x = 1,2,3.

    y = 2,4,6.

    three numbers 12,24 and 36.

    Divisibility rules in various Base system

    In any base system 'n' a number is divisible  by 'n-1' depends upon sum of digits of number.

    As we use base 10; if we have to check any number is divisible by 9,we find sum of digits of number and divide by 9.

    suppose let's check 297 is divisible by 9 or not.

    2 + 9 + 7 = 18.

    we get digit sum = 18.

    18 mod 9 = 0.

    297 is divisible by 9.

    Find the remainder when (8484)_12 is divided by 11.

    8 + 4 + 8 + 4 = 24.

    24 mod 11 = 2.

    Find remainder when (99987)_11 divided by 10.

    9 + 9 + 9 + 8 + 7 = 42.

    42 mod 10 = 2.

    In any base n, any number divided by (n + 1) leaves the remainder as same as the difference between sum of the digits of number in odd places and that in even places is divided by (n + 1).

    Let's check 13486 is divisible by 11 or not.

    (1 + 4 + 6) - (3 + 8 ) = 0.

    divisible by 11.

    Find the remainder (8484)_12 is divided by 13.

    (4 + 4) - (8 + 8 ) = -8.

    -8+13 = 5.

    remainder = 5.

    find the remainder (76456)_8 divided by 9.

    (7 + 4 + 6) - ( 6 + 5) = 6.

    remainder = 6.

    In any base n, a number divided by a factor by factor of n leaves same remainder as the last digit of number is divided by factor of n.

    suppose in base 10, any no is divisible by 2 & 5.

    Then last digit of both number both should be divisible by 2 & 5.

    To check any no is divided by 4, we check last 2 digits are divisible by 4 or not. ( 4 = 2^2).

    To check any no is divided by 8, we check last 3 digits are divisible by 8 or not.

    Q. A 99 digit number is formed by writing first 54 natural numbers in front of each other as 123 ... 54.

    Find the remainder when this number is divided by 8. ( CAT 1998 ).

    (a) 4                    (b) 7

    (c) 2                    (d) 0

    Solution :-

    We know 2 is factor of 10.

    8 = 2^3.

    Hence 8 must be factor of 10^3.

    We need to divide last 3 digits only.

    354  mod 8 = 2.  



  • Sir, please explain the last question again.


  • NMIMS, Mumbai (Gold Medallist) | Asia-Pacific Champion – CFA Global Research Challenge


    @Swami-G

    For the divisibility by 8 , we check only for the last three digits of the number. If the last three digits of a number is divisible by 8, then the number will be divisible otherwise not. For the remainder , the remainder left by the division of last three digits by 8 will be the remainder when the whole number is divided by 8.
    So, here the number is 1234567891011121314…….525354.
    Last three digits of the number = 354
    Remainder when the number is divided by 8 = Rem(354/8) = 2


  • NMIMS, Mumbai (Gold Medallist) | Asia-Pacific Champion – CFA Global Research Challenge


    @vikas_saini Just curious, why the last question is added in a Base System chapter. Seems more like a Divisibility rule concept than base system.

    Your articles are very good. Thanks for all the effort :thumbsup:



  • @harris Thank you very much sir..it was old cat question, so I added it.


  • NMIMS, Mumbai (Gold Medallist) | Asia-Pacific Champion – CFA Global Research Challenge


    @vikas_saini Good. Just asked :slight_smile:


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