For a natural number $n$ say that $d(n)$ is the sum of the digits of $n$ (in base $10$). Then what is the value of $$d(d(d(4444^{4444}))) ?$$ I have been trying with modular arithmetic, but can't do it.
Asked
Active
Viewed 448 times
6
-
Thinking about that problem makes my head spin. – Gregory Grant Mar 11 '15 at 15:07
-
It may help to notice that $d(4444^2)=40$. – barak manos Mar 11 '15 at 15:08
-
The answer is supposed to be $7$; on the first iteration, it is $72601$; on the second, $16$. – heropup Mar 11 '15 at 15:11
-
If you search MSE under 4444, you will find the question has been asked several times. – André Nicolas Mar 11 '15 at 15:12
1 Answers
4
Note that the sum of digits function is essentially a logarithm-for a reasonable mix of digits $d(n)\approx 4.5 \log_{10}(n)$ Also, the sum of digits maintains the value $\pmod 9$. So if you can compute $4444^{4444} \pmod 9$, then convince yourself that $d(d(d(4444^{4444}))) \lt 10$ you are home.
Ross Millikan
- 374,822
-
-
It is $4444^{4444}\mod 9$. As $4444\equiv 7\mod 9$ and $7$ has order $3$ modulo $9$, we have that: $$4444^{4444}\equiv 7^{4444}=7^{4444\mod 3}=7^1=7\mod 9.$$ – Bernard Mar 11 '15 at 15:20
-
@gammatester: yes, you have to use $9$ as the value for multiples of $9$, not $0$ – Ross Millikan Mar 11 '15 at 15:22