I see my grand total is not the same as Charlie's. We only differ by 1. I assume I overcounted by 1 somehow.
For changing two numbers at a time I considered subcases separately, being careful not to overcount from Case 2 in my other post.
Case 3a. Changing two numbers in odd positions. (10 pairs)
A +8 or -3
Of these pairs, most gave 7 possibilities, only the pair (8,4) gave 8.
Total _71_
Case 3b. Changing two numbers in even positions. (10 pairs)
B -8 or +3
Of these pairs, five gave 7 and five gave 8
Total _75_
Case 3c. Changing one of each (25 pairs)
|A-B| -3 or +8 or -14
Of these pairs, seventeen have 7 and eight gave 8
Total _183_
For a grand total 9+9+71+75+183=347.
(Charlie gives 346)
Sample for case 3a. The pair (1,8) sum is 9. Change it to 9+8=17 or 9-3=6.
The 7 pairs with this sum, changing both digits are
(8,9) (0,6) (2,4) (3,3) (4,2) (5,1) (6,0) Don't count (9,8) or (1,5)
82379945560
02376945560
22374945560
32373945560
42372945560
52371945560
62370945560
Sample for case 3c. The pair (8,7) difference is 1. Change it to 1-3=-2 or 1+8=9.
There are nine pairs with these differences.
Count (0,2)(1,3)(2,4)(3,5)(4,6)(6,8)(7,9)(9,0) but not (5,7)
12320945560
12331945560
12342945560
12353945560
12364945560
12386945560
12397945560
12309945560
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Posted by Jer
on 2021-02-08 10:47:57 |
Harder solutions, full answer
