Problem 1084. Square Digits Number Chain Terminal Value (Inspired by Project Euler Problem 92)
Let consider the case x=954
954 -> 122 -> 9 -> 81 -> 65 -> 61 -> 37 -> 58 -> 89 -> 145 -> 42 -> 20 -> 4 -> 16 -> 37
So 37 is seen twice, it should end up with 37, no ?
No, 89 shows up before 37 does
By the description at ProjectEuler.net, it seems M is correct in their statement. The description there reads: "continuously [...] form a new number until it [that number just formed] has been seen before." That is exactly what M has done. Project Euler says furthermore: "[...] EVERY starting number will eventually arrive at 1 or 89." That is _not_ the same as stating: "this number chain always terminates with either 1 or 89". The key difference is in "arrives at" (the number appears) versus "terminates with" (the number is the first to appear twice in the sequence). In M's example, the sequence truly 'terminates with' 37 (as M said), but before it terminates the sequence had 'arrived at' 89.
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