I am told this may be connected to the properties of numbers modulo 9.
Thank you.
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7You could look here http://math.stackexchange.com/questions/169797/write-down-the-sum-of-sum-of-sum-of-digits-of-44444444 – Frank Oct 31 '16 at 17:15
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This is the most famous IMO problem I think – Asinomás Oct 31 '16 at 17:16
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1Do you mean, performing "sum of digits" until reaching a single digit? Or do you mean, performing "sum of digits" $3$ times? – barak manos Oct 31 '16 at 17:18
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Yes, repeatedly summing the digits of a positive integer N eventually yields the residue class of N mod 9, except that when N is divisible by 9 the final sum of digits is 9 rather than 0.
So the first task is is compute the class of $4444^{4444}$ mod 9.
When computing the class of a power $r^s$ mod 9, note that the 6th power of r is 1 mod 9 provided that r is not divisible by 3. This allows us to reduce s modulo 6 before we begin. 4444 is not divisible by 3, and 4444 is 4 mod 6 so $4444^{4444} \equiv 4444^4$ mod 9.
To compute this class mod 9, we can begin by reducing that remaining 4444 mod 9. 4444 is 7 mod 9 so $4444^{4444}\equiv 7^4$ mod 9.
$7^4$ is 49 times 49, and 49 is 4 mod 9. We have $7^4 \equiv 4^2$ mod 9, and $4^2$ is 7 mod 9.
So $4444^{4444}\equiv 7$ mod 9, and the sum of digits you seek is 7
DCarter
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It's also necessary to verify that the process terminates after three steps, e.g. using logs. – Andrew Dudzik Oct 31 '16 at 17:56