Let $n$ be a positive integer. We will denote by $\sigma(n)$ its sum of digits in base 10. It is very easy to show that $\sigma(n) \leq 9(\log n + 1).$ So, $A = \sigma(4444^{4444}) \leq 9(\log(4444^{4444})+1) \leq 9(4444 \log(10^5)+1) = 199989.$ Then, $B = \sigma(A) \leq 46,$ then $\sigma(B) \leq 12.$ So, $1 \leq \sigma(B) \leq 12.$ Besides, we know that modulo 9 : $\sigma(b) \equiv B \equiv A \equiv 4444^{4444}.$ Finally, $4444^{4444} = (9 \cdot 443 + 7)^{4444} \equiv 7^{4444} \pmod 9.$ Moreover, $7^3 \equiv 1 \pmod 9$ and so $7^{4444} \equiv 7 \pmod 9.$ Eventually, $1 \leq \sigma(B) \leq 12$ and $\sigma(B) \equiv 7 \pmod 9$ which implies $\sigma(B) = 7.$

We have $ 4444^{4444}\leq10000^{5000}$ so $ 4444^{4444}$ has fewer than $ 20,000$ digits. So $ A$ is less than $ 180,000$. Among the numbers less than $ 180,000$ the number with the largest sum of digits is $ 179,999$ with the sum $ 44$. So $ B\leq44$, and the sum of the digits of $ B$ is at most the sum of the digits of $ 39$ which is $ 12$. But every number is congruent to the sum of its digits modulo $ 9$. Hence $ 4444$ is congruent to $ 7$, modulo $ 9$ and we have $ 7^{3}\equiv7(mod 9)$. It follows that $ 4444^{4444}$ is congruent to $ 7$ modulo $ 9$. Hence the answer is $ 7$.

manjil wrote: and we have $ 7^{3}\equiv7(mod 9)$. No, we don't.

Approach by ZetaX: when $ Q(n)$ is the digit sum: $ n = \sum n_i 10^i \equiv \sum n_i = Q(n) \mod 9$ So we only have to find $ 4444^{4444} \mod 9$, but $ 4444 = 9 \cdot 494 - 2 \equiv - 2\mod 9$, so: $ 4444^{4444} \equiv ( - 2)^{4444} \mod 9$. But now as said in the script, $ ( - 2)^3 \equiv - 8 \equiv 1 \mod 9$ and so $ ( - 2)^{4444} = ( - 2)^{3 \cdot 1481 + 1} \equiv ( - 2)^1 \equiv 7 \mod 9$, and its done. We know that $ B \leq 54$ and that $ B = Q(A) \equiv A = Q( 4444^{4444} ) \equiv 4444^{4444} \equiv 7 \mod 9$. But there are only finitly many numbers with that property, and writing down all of them (namely $ B = 7,16,25,34,43,52$) we see that all of them fulfill $ Q(B) = 7$, so there is no other possibility left.

Sorry I wanted to mean $ 7^{3}$ is congruent to $ 1$ modulo $ 9$. That was a typo!

Solution: $d(n) \equiv$ sum of digits of $n$. We want $d(d(d(4444^{4444})))$ Note: 1. $d(\#\#) \leq d(99) = 18$.(\# is denoting digits) 2. Warning: $m>n$ doesn't imply $d(m) > d(n)$ (Not always.) Example: $1111>7$ but $d(1111)=4<d(7)=7$. 3. We know that $100<125<1000$, so, if we apply $\log_{10}$, (note that the inequality will maintain, as $\log$ is an increasing function) we will get: $2<\text{log}_{10}125<3$. It basically tells us (about) the number of digits of a number. Check: $\log_{10}(4444^{4444}) = 4444 \times \log_{10}4444 < 4444 \times 4 =17776$. So, $4444^{4444}$ will have no more than $17776$ digits. We are looking for an upper bound on the digits' sum. Now we can say \[ d(4444^{4444}) < \overbrace{999...9}^\text{17776 9's} = 9 \times 17776 = 159984\]We have $d(4444^{4444})<159984$. So,$$d(d(4444^{4444}))<d(199999)=46$$$$\implies d(d(d(4444^{4444}))) < d(49)=13$$Question. Why are we keeping the leftmost digit of the numbers in the RHS of the inequalities intact? Ans: Let us take an example, $46>39$ but $d(46)=10 \not >d(39)=12$. So, by keeping the leftmost digit intact we are ensuring that the (upper) bound on the digit sum is correctly maintained and properly established. And secondly, we are also establishing a sharper bound on the digit sum, by doing so; as $9$ is the largest digit in the set of digits $(0,1,2...9)$, so, without replacing the leftmost digit by $9$ (which may be correct, but then it would make the digit sum larger) we are keeping it intact and replacing the following digits with $9$. So, $$\boxed{d(d(d(4444^{4444}))) < 13} -(*)$$FACT: $n \equiv d(n) \ (mod\ 9)$ Note that $4444 \equiv 7 \ (mod\ 9)$. $\implies 4444^2 \equiv 7^2 \equiv 4 \ (\text{mod}\ 9) \implies 4444^4 \equiv 4^2 \equiv 7 \ (\text{mod}\ 9)$ $\implies 4444^8 \equiv 7^2 \equiv 4 \ (\text{mod}\ 9) \implies 4444^{16} \equiv 7\ (\text{mod}\ 9)$ $\implies 4444^{32} \equiv 4\ (\text{mod}\ 9) \implies 4444^{64} \equiv 7\ (\text{mod}\ 9)$ $\implies 4444^{128} \equiv 4\ (\text{mod}\ 9) \implies 4444^{256} \equiv 7\ (\text{mod}\ 9)$ $\implies 4444^{512} \equiv 4\ (\text{mod}\ 9) \implies 4444^{1024} \equiv 7\ (\text{mod}\ 9)$ $\implies 4444^{2048} \equiv 4\ (\text{mod}\ 9) \implies 4444^{4096} \equiv 7\ (\text{mod}\ 9)$ So, $$4444^{4444} \equiv 4444^{4096} \times 4444^{256} \times 4444^{64} \times 4444^{16} \times 4444^{8} \times 4444^{4} \equiv 7 \cdot 7 \cdot 7 \cdot 7 \cdot 4 \cdot 7\ (\text{mod}\ 9) $$$$\boxed{4444^{4444} \equiv 7 (\text{mod}\ 9)}$$$$\implies d(4444^{4444}) \equiv 7 (\text{mod}\ 9) \implies d(d(4444^{4444})) \equiv 7 (\text{mod}\ 9)$$$$ \implies \boxed{d(d(d(4444^{4444}))) \equiv 7 (\text{mod}\ 9)}-(**)$$By $(*)$ and $(**)$ we have $$\boxed {d(d(d(4444^{4444}))) = 7}$$Verification: According to Wolfram Alpha, the digit sum of $4444^{4444}$ is $72601$. $\implies d(4444^{4444})=72601 \implies d(72601) = 16 \implies d(16)=7$.

here we need to compute: s(s(s(4444^{4444}))), using the inequality : s(n) \le 9([logn]+1) several times, we have : s(4444^{4444})\le 9([log 4444^{4444}] +1)<9\times20000=180000 now , again we have : s(s(4444^{4444}))\le 9([log s(4444^{4444})] +1)\le9(log 180000 +1)\le 63. so, s(s(s(4444^{4444})))\le 14 (as among all the numbers 1 to 63 , maximum value of sum of digits is 14) on the other hand , s(s(s(n)))\equivs(n)\pmod{9} again , s(n)\equiv n\pmod{9} and since 4444^{4444}\eqviv7^{4444}\eqviv 7\times 7^{3\times1481\equiv 7\pmod{9}

Let $X = 4444^{4444}$. Then, $X$ has at most $17776$ digits as $$\log X + 1 = 4444 \log 4444 + 1 < 4444\cdot 4 + 1 = 17777.$$Therefore, $S(X) \leq 9 \cdot 17776 = 159984$ and $S(S(X)) \leq S(99999) = 45$. Finally, $S(S(S(X))) \leq S(39) = 12$. Additionally, since $S(S(S(X))) \equiv X \equiv 7 \pmod{9}$, we must have $S^3(X) = \boxed{7}$.

Let $S(n)$ denote the sum of the digits of $n$, so we want $S(S(S(4444^{4444}))) = S(S(A))= S(B)$. Claim 1: $S(n) \equiv n$ mod $9$ for all $n$ Let $$n = \overline{a_k} \; \overline{a_{k-1}} \cdots \overline{a_1} = 10^{k-1} \cdot a_k + 10^{k-2} \cdot a_{k-1} + \cdots + 10^0 \cdot a_1.$$ Then, when taken mod $9$, we get that $$n \equiv 1^{k-1} \cdot a_k + 1^{k-2} \cdot a_{k-1} + \cdots + 1^0 \cdot a_1 \equiv a_k + a_{k-1} + \cdots + a_1 = S(n),$$as desired. Claim 2.1: $0 < A \le 148005$ Trivially $0 < A$. For some arbitrary $n$ with $\mathcal P$ digits, $S(n)$ is maximized when $\underbrace{n = 999\cdots999}_{\mathcal P ~ 9\text{s}}$, so if $n$ has $\mathcal P$ digits, then $S(n) < 9 \cdot \mathcal P$. This also means that $S(n)$ is maximized when $n$ has as many digits as possible. Note that $4444^{4444} < \dfrac{10000^{4444}}{2^{4444}}$. If we find a lower bound for $2^{4444}$, we can thus find an upper bound for $4444^{4444}$. Evidently $2^{4444} > 2^{4440}$. Since $2^{10} > 10^3$, we get that $2^{4440} > 10^{444 \cdot 3}$, so $2^{4444} > 10^{1332}$. Thus, $4444^{4444} < \dfrac{10000^{4444}}{2^{4444}} < \dfrac{10000^{4444}}{10^{1332}}$, so thus it has at most $16445$ digits and $S(4444^{4444}) \le 16445 \cdot 9$, so thus $A \le 148005$, as desired. Claim 2.2: $0 < B \le 45$ Trivially $0 < B$. We have that $A \le 148005$. If $A$ has $5$ digits, then $B$ is at most $45$. If $A$ has $6$ digits, then the first digit of $A$ will increase by $1$ (from $99999$), but then its second digit will decrease by more than $1$ (from $99999$), so this will not give a higher upper bound for $B$. Thus, $B \le 45$, as desired. Claim 2.3: $0 < S(B) \le 12$ Trivially $0 < S(B)$. Since $B \le 45$, we get that $S(B)$ is at most $12$ (for $B = 39$), as desired. Claim 3: $4444^{4444} \equiv 7$ mod $9$ Note that $4444^{4444} \equiv 7^{4444}$ mod $9$, and since $7^3 \equiv 1$ mod $9$, we get that $4444^{4444} \equiv (7^3)^{1481} \cdot 7 \equiv 1^{1481} \cdot 7 \equiv 1 \cdot 7 \equiv 7$ mod $9$, as desired. Ending: By Claim 1 and Claim 3, we get that $S(B) \equiv 7$ mod $9$, and by Claim 2.3, we get that $0 < S(B) \le 12$. Thus, $S(B) = \boxed{7}$. Remark: The bounding in Claim 2 didn't have to be that strict, but whatever.

Let $S(n)$ be the sum of the digits of $n$ and let $C=S(B)$. We know that $S(n)\equiv n\pmod9$. Then: $$A\equiv4444^{4444}\equiv(-2)^{4444}\equiv2^{4444}\equiv2\cdot8^{1481}\equiv2\cdot(-1)^{1481}\equiv-2\equiv7\pmod9$$and $$C\equiv B\equiv A\equiv7\pmod9.$$Now we bound $C$. Since $4444^{4444}\le10000^{4444}=10^{22220}$, $A\le22220\cdot9=199980$. Since $A\le10^7$, $B\le7\cdot9=63$ and $C\le6+9=15$. But in order for $C$ to be $7\pmod9$, it must be exactly $\boxed7$.

Use a program n = 4444Preon = n ** nPreon = str(Preon)s = 0for i in range(len(Preon)): s += int(Preon[i])s = str(s)s1 = 0for i in range(len(s)): s1 += int(s[i])s2 = 0s1 = str(s1)for i in range(len(s1)): s2 += int(s1[i])print(s2)n = 4444 Preon = n ** n Preon = str(Preon) s = 0 for i in range(len(Preon)): s += int(Preon[i]) s = str(s) s1 = 0 for i in range(len(s)): s1 += int(s[i]) s2 = 0 s1 = str(s1) for i in range(len(s1)): s2 += int(s1[i]) print(s2)RunResetPop Out /

Simply use the bound $$S(n) \leq 9(\log_{10}(n) + 1).$$Observe that $$S(4444^{4444}) \leq 9(\log_{10}(4444^{4444}) + 1) < 9(4444 \cdot 4 + 1) = 159993.$$Thus $$S(S(4444^{4444})) \leq S(99999) = 35.$$All we will need is $$S(S(S(4444^{4444}))) < 16,$$which is evident. Now notice that $S(n) \equiv n \pmod 9$, and one can check that $4444^{4444} \equiv 7 \pmod 9$. Thus, $$S(S(S(n))) \equiv 7 \pmod 9,$$and by our earlier inequality it must be precisely $\boxed 7$.