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I have this problem:

Compute $ sd(sd(sd(1976^{1976}))) $, where $sd()$ represents the sum of digits.


I know that $1976 = 2^3 \cdot 19 \cdot 13$, so we have to compute $ sd(sd(sd((2^3 \cdot 19 \cdot 13)^{1976}))) $.

I also know that $sd(a\cdot b) = sd(sd(a)\cdot sd(b))$, so I can rewrite the above as: $sd(sd(sd( sd(2^{3\cdot1976}) \cdot sd(13^{1976}) \cdot sd(19^{1976})))$ but I think this is not the right direction.

I also know that $sd(m\cdot n) \leq sd(m)\cdot sd(n)$, so theoretically I should be able, maybe, to approximate the sum, but how to compute the sum of sum of sum of digits of that number ($1976^{1976})$?

Thank you very much!

Note: I previously asked how to compute $sd(1976^{1976})$ but it seems there was a typo in my book and the real question is the question I ask here. I edited the old question, to the iterative sum and accepted the correct answer.

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    $\begingroup$ @JoséCarlosSantos I do think the problem is slightly different. $\endgroup$ – Don Thousand Oct 22 '18 at 15:42
  • $\begingroup$ According to WA the sum of digits of $1976^{1976}$ is $29239$ and the sum of digits of that is (obviously) $25$ so the next one up is $7$. This suggests: try to prove that your value is just the iterated sum of digits. $\endgroup$ – lulu Oct 22 '18 at 15:43
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It is easy to see that the iterated sum of digits of your expression is $7$, thus your answer must be congruent to $ 7 \pmod 9$.

To solve the given problem:

Note that $$\log_{10}{1976^{1976}}\approx 6512.474994$$ so your expression has $6513$ digits. Thus the sum of digits of your expression is not greater than $$9\times 6513=58617$$

Now, that has $5$ digits so the second sum of digits is less than $9\times 5=45$. Now, inspection quickly shows that there is no number $≤45$ which has a sum of digits greater than $12$. Thus the answer to your question is also $7$.

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  • $\begingroup$ Thank you very very much!!!! $\endgroup$ – MM PP Oct 22 '18 at 17:41
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You need to calculate $1976^{1976}\mod {9}$. Be decomposing $1976$ as you did we have $$1976^{1976}=(2^3.19.13)^{1976}=(4.19.26)^{1976}$$as we know $$26\cong -1\mod 9\\19\cong 1\mod 9$$therefore$$1976^{1976}\cong 2^{2\times 1976}\mod 9\cong 2^{2\times 1974}\cdot 16\cong 7\mod 9$$so the sum of the digits after too many levels is $7$

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