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Sum of all possible Numbers with Repetition

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Sum of all possible Numbers with Repetition

by ronnie1985 » Wed May 23, 2012 8:22 am
What is the sum of all 4-digit numbers that can be formed using the digits 1,2,3,4 where repetition of digits is allowed?

What's the approach? We know that 256 such numbers are possible.
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by coolhabhi » Wed May 23, 2012 11:08 am
Since each number is repeated they will appear equal number of times.

the total numbers that can be formed are 4^4 = 256 numbers..

Of these numbers the units digit will be taken by 1 in (256/4) = 64(Since total numbers are 256 and the digits 1,2,3,4 are four in number).

Similarly the units digit will be taken by 2 in (256/4) = 64
Similarly the units digit will be taken by 3 in (256/4) = 64
Similarly the units digit will be taken by 4 in (256/4) = 64

So the sum of digits in units place is 64(1 + 2 + 3 + 4) = 640

the same way the sum of digits in tens place is 640(1 + 2 + 3 + 4) = 6400
the sum of digits in hundreds place is 6400(1 + 2 + 3 + 4) = 64000
the sum of digits in thousands place is 64000(1 + 2 + 3 + 4) = 640000

The total is 640000 + 64000 + 6400 + 640 = 711040. :D
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