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I found an interesting exercise on Number Theory (maybe interesting just for me, as I don't know how to solve it).

Compute the iterative sum of digits of: $1976^{1976}$.

I really don't know how to solve this exercise. I noted that $2025$ is the next perfect square and $2025-1976=49$, so $1976=2025-49$, and $49$ is a perfect square too. So I have to compute the sum of digits of $(45^2 - 7^2)^{45^2-7^2}$. I don't know how would that help me, but seemed like a hint to me when I found these two square roots.

Please help me. Thank you very much!

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    $\begingroup$ @Arthur I did not said that the difference between two perfect squares is a square, but that both numbers, 2025 and 49 are perfect squares. Thank you! $\endgroup$ – MM PP Oct 22 '18 at 14:32
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    $\begingroup$ You're right, I misread. $\endgroup$ – Arthur Oct 22 '18 at 14:32
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    $\begingroup$ Are you looking for the sum of the digits or for the iterated sum of the digits (which is essentially just the residue $\pmod 9$)? I don't really see how to get the literal sum of digits without brute force. To be sure, it's not that hard to do with brute force (just a couple of seconds in WA). $\endgroup$ – lulu Oct 22 '18 at 14:48
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    $\begingroup$ Please tell us where you found this exercise. I do not think there is any shortcut to this problem, as stated, other than explicit computation. $\endgroup$ – астон вілла олоф мэллбэрг Oct 22 '18 at 14:51
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    $\begingroup$ I think this is from some 1976 mathematical olympiad, and the correct version is: Compute the sum of the sum of the sum of the digits of $1976^{1976}$. But I can't find it online. It's not IMO 1976. $\endgroup$ – TonyK Oct 22 '18 at 15:03
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(I am assuming "sum of digits" means "iterated sum of digits")

First note that the iterated sum of the (decimal) digits of a number $n$ is equal to $n \mod 9$.

Then note that $(a^b) \mod 9= ((a \mod 9)^b) \mod 9$.

So

$(1976^{1976}) \mod 9 = ((1976 \mod 9)^{1976}) \mod 9 = (5^{1976}) \mod 9$

Now calculate the first few powers of $5$ modulo $9$:

$5^2 \mod 9 = 25 \mod 9 = 7$

$5^3 \mod 9 = 5 \times 7 \mod 9 = 35 \mod 9 = 8$

$5^4 \mod 9 = 5 \times 8 \mod 9 = 40 \mod 9 = 4$

$5^5 \mod 9 = 5 \times 4 \mod 9 = 20 \mod 9 = 2$

$5^6 \mod 9 = 5 \times 2 \mod 9 = 10 \mod 9 = 1$

$5^7 \mod 9 = 5 \times 1 \mod 9 = 5 \mod 9 = 5$

Can you see you can use this pattern to find $(5^{1976}) \mod 9$ ?

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