Find the
digit sum
of the sum of all the
numbers that can be formed using
without repetition.
Note : Digit sum of "12345" is 1+2+3+4+5=
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Let us think about the combinations in ways of arranging the digits. There would be 9 ! ways to arrange any digit at any given place. Therefore there would be 9 ! of each 1 , 2 , 3 . . . 9 (Zero doesn't contribute to the final digit sum.)
So the total value at each given place is 9 ! × ( 1 + 2 + 3 . . . . + 8 + 9 ) , and these can be arranged in any of the 10 places (units to billions place).
So the sum of all the 10-digit numbers would be : 9 ! × 4 5 × ( 1 + 1 0 + . . . . + 1 0 0 0 0 0 0 0 0 0 ) 9 ! × 4 5 × ( 1 1 1 1 1 1 1 1 1 1 ) 1 8 1 4 3 9 9 9 9 9 8 1 8 5 6 0 0 (Yes, I used a calculator for the above step, I don't know if there's an easier way to get to the direct digit sum rather than calculate it )
The digit sum of which is : 9 0