Mission IIMpossible - Gyan Room (Modern Math - Short cuts/Questions/Concepts)



  • Gyan Room - There's always room to learn more!
    This thread is reserved for sharing concepts, short cuts and good questions from Modern Math topic.

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    Rules:

    1. Mention source (if required)
    2. Provide examples for short-cuts
    3. Posts not related to Modern Math will be removed.

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


    Sum of all numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time (means repetition is not allowed)

    The total number of numbers formed with the digits 2, 3, 4, 5 taken all at a time = 4! = 24
    to find the sum of these 24 numbers we will find sum of digits at unit, ten, hundred and thousand places in all these numbers.
    consider the digits in the unit's places in all these numbers
    Each of the digit will occur 3! =6 times in the unit place
    So total for the digits in unit's place in all the numbers =(2 + 3 + 4 + 5) * 6
    so sum of all numbers =(2 + 3 + 4 + 5) * 3! * (10^0+10^1+10^2+10^3))
    so (2 + 3 + 4 + 5) * 3! * (10^4-1)/9
    = (2 + 3 + 4 + 5) * 3! * (1111)

    Direct Formula:
    Sum of numbers formed by n non zero digits = (sum of digits) * (n-1!) * (11111....n times)

    When repetition allowed
    total number of numbers in case of repetition = 4 * 4 * 4 * 4 = 4^4
    So each digit will occur at 4^4/4 = 4^3 = 64 times
    so (sum of digits) * 64 * (1111))

    Direct Formula = sum of digits * n^n * (1111)


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


    Find sum of all the numbers greater than 10000 formed by digits 0, 2, 4, 6, 8 (No digit is repeated in any number)

    Sum = sum of numbers formed by using digits (0,2,4,6,8) - sum of numbers formed by using digits (2,4,6,8)
    (0 + 2 + 4 + 6 + 8) * (5-1!) * (11111) - (2 + 4 + 6 + 8) * (4 - 1!) * (1111))


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