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Extreme Sum Settlement (Posted on 2015-09-28) Difficulty: 3 of 5
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    
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Solution Solution (barring errors) | Comment 1 of 4
Sum | ways | sum-15 | ways | product  
15 519 0 1 519
16 564 1 4 2256
17 597 2 10 5970
18 615 3 20 12300
19 615 4 35 21525
20 597 5 56 33432
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35 4 20 633 2532
36 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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