Find the sum of all of the three digit positive integers with three distinct digits.
Here's an analytic strategy that calculates the average of each digit in a 3digit number with unique digits.
There are 9 ways to choose the first digit, 9 ways to choose the second, and 8 ways to choose the third. The number of 3digit numbers with unique digits is 9*9*8 = 648.
The average first digit (which cannot be zero) is 5, whereas the other digits' average value is less than that. If the choosing of digits 2 and 3 were independent of the first digit, then the average value of digits 2 and 3 would be 4.5. Then the answer would be 648 * (100*5 + 10*4.5 + 4.5) = 356076.
But the value of digits 2 and 3 is not independent of the value of digit 1; it is biased slightly towards zero since digit 1 can never be zero.
Consider if the first digit were 1, then digits 2 and 3 would have an average value of (0+2+3+4+5+6+7+8+9)/9. If you repeat that process nine times, each time eliminating one digit from (1,2,...,9), you get an average value for digits 2 and 3 of 40/9.
So the equation becomes
648 * (100*5 + 10*40/9 + 40/9) = 355680

Posted by Larry
on 20240328 09:43:23 