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hemant_malhotra last edited by
Director at ElitesGrid  CAT 2016  QA : 99.94, LRDI  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! * (10^41)/9
= (2 + 3 + 4 + 5) * 3! * (1111)Direct Formula:
Sum of numbers formed by n non zero digits = (sum of digits) * (n1!) * (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)

hemant_malhotra last edited by
Director at ElitesGrid  CAT 2016  QA : 99.94, LRDI  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) * (51!) * (11111)  (2 + 4 + 6 + 8) * (4  1!) * (1111))