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weekcountingproblem [2026/02/13 02:30] – [Possible Solutions] thehomelandweekcountingproblem [2026/02/13 02:42] (current) – [Possible Solutions] thehomeland
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 ^ 38 counts ^ 42 counts ^ 46 counts ^ 51 counts ^ 52 counts ^ ^ 38 counts ^ 42 counts ^ 46 counts ^ 51 counts ^ 52 counts ^
-| 8m1w | 9m1w | 10m1w | 11m1w | 12m1w |+| 8m1w | 9m1w | 10m1w | 11m1w | 1y1w |
 | 8m2w | 9m2w | 10m2w | 11m2w |  | | 8m2w | 9m2w | 10m2w | 11m2w |  |
 | 8m3w | 9m3w | 10m3w | 11m3w |  | | 8m3w | 9m3w | 10m3w | 11m3w |  |
-| 9m | 10m | 11m | 12m |   | +| 9m | 10m | 11m | 1y |   | 
-|   |    |   | 12m0w |   |+|   |    |   | 1y0w |   |
  
-This seems like a viable solution, but might be difficult to remember, especially when lots of people blow right past it messing up the count.  Is there another more foolproof option, aside from ignoring it? +This seems like a viable solution, but wouldn't that therefore make the start of the next count of 1y2w be the starting count for the next 52, whereas the start of the first 52 was 0m0w? 
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