A recent question asked about the sum of sum of sum of digits of $4444^{4444}$. The solution there works mainly because the number chosen is small enough for the sum of sum of sum to be equal to the repeated sum: i.e. if we sum digits further, the result does not change. Since finding repeated sums of digits is just a matter of elementary number theory, this solves the problem.

It seems the following question might be much harder: what is the sum of sum of digits of $$4444^{4444^{4444}}?$$ In other words, let $f:\Bbb N_0\to\Bbb N_0$ be the function defined by $f(n)=\textrm{sum of decimal digits of }n$.

What is the value of $f\left(f\left(4444^{4444^{4444}}\right)\right)$?

In this question, we have not yet reached a single-digit number, which at least seems to make it much harder.

Some estimates: the number of decimal digits of $4444^{4444^{4444}}$ is equal to $$\left\lfloor\log_{10}4444^{4444^{4444}}\right\rfloor+1,$$ which implies $$f\left(4444^{4444^{4444}}\right)\le9\left(\log_{10}4444^{4444^{4444}}+1\right).$$

Next, the number of digits of this last number is at most $$\left\lfloor\log_{10}\left(9\left(\log_{10}4444^{4444^{4444}}+1\right)\right)\right\rfloor+1,$$ which is $16213$, according to Wolfram|Alpha. Therefore, $$f\left(f\left(4444^{4444^{4444}}\right)\right)\leq9\cdot16213=145917.$$

So the number we are looking for has at most $6$ digits. This makes it very feasible to express in decimal notation, but possibly hard to find.

We might be further interested in numbers like $$f\left(f\left(f\left(4444^{4444^{4444^{4444}}}\right)\right)\right),$$ so a related question would be:

Is there any hope for a general method of evaluating such functions or is the behaviour of the $k$-fold composition $f^k$ completely chaotic?

  • $\begingroup$ smells like an advanced application of the chinese remainder theorem. $\endgroup$
    – akkkk
    Commented Jul 12, 2012 at 10:08
  • 7
    $\begingroup$ @Auke: May I ask, which moduli do you smell? $\endgroup$ Commented Jul 12, 2012 at 10:34
  • 1
    $\begingroup$ @MarcvanLeeuwen: mod 10, mod 100, mod 1000, etc. $\endgroup$
    – akkkk
    Commented Jul 12, 2012 at 11:17
  • 9
    $\begingroup$ @Anke: The Chinese remainder theorem doesn't work very well for moduli that are multiples of each other. $\endgroup$ Commented Jul 12, 2012 at 13:24
  • 1
    $\begingroup$ Have you seen this ? I think one can use the same thread of idea and extend it recursively to a higher notion. $\endgroup$
    – IDOK
    Commented Jul 21, 2012 at 8:49

1 Answer 1


You can find an upper bound for it even without using computer or any calculator: $$ f(N) < 9 (4444^{4444} \times \log_{10} 4444 + 1) < 9 \times 4 \times 4444^{4444} + 9 $$ $$ f(f(N)) < 9 ( \log_{10}9 + \log_{10}4 + 4444 \log_{10}4444 + 1) < 9 (3 + 4444 \times 4) = 9 \times 17779 = 160011 $$

so $$ f(f(N))<160011 $$ this is a large range but it can be smaller with calculator. (Note that you should have computed base 10 logarithm instead of natural logarithm)

the range consists of 160011 numbers, and by knowing the reminder of 9, only 17,779 numbers are left, and answer is one of them.

Of course this is not an exact answer, but it is straightforward!

Edit: I just misused a formula which is about something different, sorry for that!

  • $\begingroup$ Thanks for the correction, I edited my question. I'm not sure I understand your lower bound, though. Where did you get $f(N)> 9(4444^{4444} + log_{10}4444 -2) > 9 \times 3 \times 4444^{4444} - 18$ from? $\endgroup$
    – Dejan Govc
    Commented Aug 11, 2012 at 23:26
  • $\begingroup$ As I'm looking at it again, I see that I just misused something I had read somewhere (which is about probability and digit count of a normal number will be around that) and even, what I have used is not that formula, so I was totally wrong, sorry about that! $\endgroup$
    – PLuS
    Commented Aug 12, 2012 at 16:26

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