For how many three-digit positive integers is the sum of the digits equal to \(19\)
+130071
9 9 1 = 3 8 9 2 = 6 7 9 3 = 6
9 8 2 = 6 8 8 3 = 3 7 8 4 = 6
9 7 3 = 6 8 7 4 = 6 7 7 5 = 3
9 6 4 = 6 8 6 5 = 6 7 6 6 = 3
9 5 5 = 3 __ __
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24 21 18 = 63
199 , 289 , 298 , 379 , 388 , 397 , 469 , 478 , 487 , 496 , 559 , 568 , 577 , 586 , 595 , 649 , 658 , 667 , 676 , 685 , 694 , 739 , 748 , 757 , 766 , 775 , 784 , 793 , 829 , 838 , 847 , 856 , 865 , 874 , 883 , 892 , 919 , 928 , 937 , 946 , 955 , 964 , 973 , 982 , 991 , Total = 45 such numbers
Note: CPhill's last 18 are already included in the permutations of the first 2 columns.
