If $d(n)$ is the sum of digits of n, find $d(d(d(n)))$ of $n=4444^{4444}$

Asked Apr 08 '14 at 07:27

Active Apr 08 '14 at 07:27

Viewed 132 times

If $d(n)$ is the sum of digits of n, find $d(d(d(n)))$ of $n=4444^{4444}$.

My attempt:

$4444^{4444}<10000^{4444}$

Now, $\max d(10000^{4444})<9\times 17776$

Again, $\max d(159984)\le 45$

Again $\max d(45)\le 12$

So, $\max d(d(d(n)))\le 12$.

Now, what am I supposed to do? Please help.

asked Apr 08 '14 at 07:27

Hawk

Well, there's this: http://www.wolframalpha.com/input/?i=sum+of+digits+of+4444%5E4444. Can you find any pattern on $d(4444^k)$ or $d(k^k)$ you might use? – Ian Coley Apr 08 '14 at 07:33
http://math.stackexchange.com/questions/169857/what-is-the-sum-of-sum-of-digits-of-444444444444 – lab bhattacharjee Apr 08 '14 at 07:36
@IanColey what is meant by the base? I am not clear about it yet... – Hawk Apr 08 '14 at 07:36
http://math.stackexchange.com/questions/169797/write-down-the-sum-of-sum-of-sum-of-digits-of-44444444 – Henry Apr 08 '14 at 07:38
1

This is IMO $1975$ Q$4$ – Mark Bennet Apr 08 '14 at 07:38
Do you know about digital roots? – Mark Bennet Apr 08 '14 at 07:39
@MarkBennet might know the matter inside, but do not know about the title of the topic, so, highly probable that I do not know about it. – Hawk Apr 08 '14 at 07:40
@MarkBennet thank you for your help... – Hawk Apr 08 '14 at 07:42
@Henry thank you for the reference. – Hawk Apr 08 '14 at 07:43
@labbhattacharjee thank you for the reference... – Hawk Apr 08 '14 at 07:43
The sum of the digits of a number is $9n+r$ for some integer $n$ where $r$ is the remainder on division by $9$. Iterating the process eventually gets you to $r$ - except multiples of $9$ go to $9$. – Mark Bennet Apr 08 '14 at 07:44
@MarkBennet Okay, thank you for the help. – Hawk Apr 08 '14 at 07:47

0 Answers0