Mathemagics by Indrajeet


  • Content & PR team - MBAtious


    Indrajeet Singh is a final year mechie who has just discovered his love for Quantitative Aptitude. He likes helping aspirants preparing for CAT and other entrance examinations with self developed fundas and theories. He started a Facebook group some time ago where aspirants practice various CAT level problems, which as of now boasts of 67k+ members and is one of the most active among all the groups out there. Also, he has put up many concepts and tricks in his website, iQuanta. He recently started his own CAT course which has more than 100 aspirants and is mostly about unconventional methods to tackle CAT Quant.

    Few of the tricks from the genius are here :

    1) Summation of the series  1/1 + 1/2 + 1/3 .... + 1/x = ?

     

    It's summation is kind of difficult to find and not known to many . So engineers can just integrate  1/x from 1 to x

     So  ∫ 1/x = ln(x)  (Log base e )

    So your summation of HP 1/1 + 1/2 + 1/3 ...+ 1/x = lnx.  

    You guys just remember the final result

    2) First digit of a number , yes first digit not last which everyone knows .

    If we need to find the first digit of a number N , then first find the value of  {logN}, let's say equal to "m" , where {x} is the fractional part of the function. Then the first digit will be given by 10^m .

    For example first digit of 3^53 :
    {Log3^53} = {53log3} = 0.28
    So first digit is 10^0.28 = 1.

    Here is how to calculate 10^0.28.
    We know 10^log2 = 2, and log2 = 0.3
    So 0.28 < 0.3, hence 10^0.28 = 1.x..
    Hence first digit = 1.

    3) Total numbers such that sum of factorial of their digits = Number itself
    Only 4 such numbers are there in the number system .
    a) 1!=1
    b) 2!=2
    c) 1!+4!+5!=145
    d) 4!+0!+5!+8!+5! = 40585

    4)  1*1! + 2*2! + 3*3+ ....n*n! = (n+1)! -1
    Many might know this but few find hard to get it's derivation
    = > (2-1)*1! + (3-1)*2! +.... (n+1- 1)n!
    = > 2*1! - 1! + 3*2! - 2! + 4*3! -3! .... (n+1)n! - n!
    = > 2! - 1! + 3! - 2! + 4! - 3! +. .. (n+1)! - n!
    Observe the pattern . So all the terms gets cancelled except first and last term So just left with  (n+1)! - 1.

     

    5) Number of primes between
    1 - 30 = > 10 primes
    1 - 50 = > 15 primes
    1 - 100 = > 25 primes
    1 - 200 = > 46 primes
    1 - 1000 = > 168 primes

    Also
    1001 = 13k ( multiple of 13)
    1003 = 17k ( multiple of 17)
    1007 = 19k ( multiple of 19)

    So 1009  is the smallest 4 digit prime.

    6) Number of positive integral solutions for a/x + b/y = 1/n  is the factors of a*b*n^2
    And integral solutions = 2 (factors of a*b*n^2) -1

    Ex : positive integral solutions of 2/x + 3/y =1/7 is factors of 2*3*7^2 = > 2*2*3 = 12
    And integral solutions is 2*12 - 1 = 23 .

    Explanation  :
     2/x + 3/y = 1/7
    = > 14y + 21x - xy = 0
    = >   y(14-x) + 21x = 0
    = > y(14-x) + 21x -14*21 = - 14*21
    = > y (14-x) - 21(14-x) = - 14*21
    = > (x-14)(y-21) = 14*21= 2*3*7^2
    = > a*b = 2*3*7^2,

    So it's just to find in how many ways the number 2*3*7^2 can be represented as product of two numbers which is = number of its factors . Hence the formula like emoticon
     
    7) Concept : A number to be written as sum of two squares :  x^2 + y^2 = N
    It will have integer solns only if N contains primes of the form 4k+1 type.
     ( it means 11, 33, 94 etc type numbers can't be represented as sum of squares but 5^3*13, 17*37^2 can be represented )
    If all primes are 4k+1 type Then positive solns = > number of factors of N,

    Number of positive solutions of x^2 + y^2 = 5^3*13^2 is 4*3 = 12
    If it contains some 4k+3 primes with odd power then number of solutions will be 0.
    If it contains even powers of 4k+3 primes, then positive solns = number of factors of 4k+1 form only.
    Integral solns = 4*positive solns.

    ( there are a bit more complications when the number is a perfect square, which you can know by joining the course )

    8 ) The number of ways of distributing 6 distinct rings in 4 fingers isn't 4^6.
    Most people do this way including few standard  text books . But r^n is wrong here .
    We apply r^n in case of balls and boxes where internal arrangement inside a box doesn't matter while in a finger the internal arrangement of ring matters and hence we can't apply r^n .
    So how to solve ? First distribute 6 rings  into 4 groups then arrange .
    a + b + c + d = 6, = > (6+4-1)C (4-1) = 9C3
    Now arrange 6 rings in 6! Ways hence your answer is 9C3*6! .


Log in to reply
 

Looks like your connection to MBAtious was lost, please wait while we try to reconnect.