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 Extreme Sum Settlement (Posted on 2015-09-28)
Determine the total number of 8-digit base ten positive integers such that the sum of the four leftmost digits is equal to precisely 15 more than the sum of four rightmost digits.

*** Assume non leading zero for each of the numbers.

 No Solution Yet Submitted by K Sengupta No Rating

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 Solution (barring errors) | Comment 1 of 4
`Sum | ways | sum-15 | ways | product  15    519     0       1          51916    564     1       4         225617    597     2       10        597018    615     3       20       1230019    615     4       35       2152520    597     5       56       33432212223242526272829303132333435      4    20       633      253236      1    21       592       592`

The first column is the total of the first four leftmost digits. The second column is the number of arrangements of digits with that sum (with no leading zeros.)  The third column is 15 less than the first.  The fourth is the number of arrangements of digits with that smaller sum (leading zeros allowed.)  The last is the product of second and fourth.

The sum of the last column is 911304 which is the solution.

I didn't feel like typing all the numbers in.  If I have made an error I may replace them.

Edited on September 28, 2015, 4:39 pm
 Posted by Jer on 2015-09-28 13:05:08

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