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April 6th, 2018, 01:09 PM   #1
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Sum of digits

Let $\displaystyle X$ be sum of digits of $\displaystyle 4444^{4444} $
Let $\displaystyle Y$ be sum of digits of $\displaystyle X$
Evaluate sum of digits of $\displaystyle Y$

Last edited by idontknow; April 6th, 2018 at 01:12 PM.
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April 6th, 2018, 03:46 PM   #2
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Ever heard of google?
https://www.google.ca/search?source=....0.KCxqWg0Hexk
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April 6th, 2018, 05:02 PM   #3
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Ever heard of google?
Well if you are just going to tell him that then why not simply use this?

I think the OP is trying to find out how to do it, not to get something to do it for him/her

Have you used modular arithmetic before? What is 4444^4444 mod 9?

-Dan
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April 6th, 2018, 05:15 PM   #4
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Suggested google as a way to learn...
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April 6th, 2018, 06:08 PM   #5
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Originally Posted by topsquark View Post
Well if you are just going to tell him that then why not simply use this?

I think the OP is trying to find out how to do it, not to get something to do it for him/her

Have you used modular arithmetic before? What is 4444^4444 mod 9?

-Dan
Ordinarily I'd agree 100% with you Dan. But not in this case. OP has a consistent history of posing difficult number theory problems out of the blue never showing a trace of work on them. It is more like they are posting brain teasers than actually asking for help.

I don't believe OP is even a student. I'm not sure what their purpose here is.
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April 6th, 2018, 08:23 PM   #6
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Ordinarily I'd agree 100% with you Dan. But not in this case. OP has a consistent history of posing difficult number theory problems out of the blue never showing a trace of work on them. It is more like they are posting brain teasers than actually asking for help.

I don't believe OP is even a student. I'm not sure what their purpose here is.
Hmph. I never noticed that about him. I'll have to keep my eyes peeled better then.

-Dan
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April 7th, 2018, 04:13 AM   #7
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Im not sure if im correct , but check below
Let $\displaystyle Z $ be sum of digits of $\displaystyle Y$
Since $\displaystyle 4444^{4444} <10000^{4444} $ number of digits of $\displaystyle 4444^{4444} $ is less then 20000
so $\displaystyle X<180000 \Rightarrow (Y,Z) \leq (45,12)$
$\displaystyle 4444 \equiv -2 (mod9)$ and $\displaystyle 2\cdot 8^{1481} \equiv 2(-1)^{1481} \equiv 7(mod9) $
$\displaystyle \Rightarrow 4444^{4444} \equiv 7(mod9)$
$\displaystyle 4444^{4444} \equiv X \equiv Y \equiv Z(mod9)$
From $\displaystyle Z \leq 12 $ and $\displaystyle Z \equiv 7(mod9)$ we have $\displaystyle Z=7$
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April 7th, 2018, 11:22 AM   #8
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Quote:
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Im not sure if im correct , but check below
Let $\displaystyle Z $ be sum of digits of $\displaystyle Y$
Since $\displaystyle 4444^{4444} <10000^{4444} $ number of digits of $\displaystyle 4444^{4444} $ is less then 20000
so $\displaystyle X<180000 \Rightarrow (Y,Z) \leq (45,12)$
$\displaystyle 4444 \equiv -2 (mod9)$ and $\displaystyle 2\cdot 8^{1481} \equiv 2(-1)^{1481} \equiv 7(mod9) $
$\displaystyle \Rightarrow 4444^{4444} \equiv 7(mod9)$
$\displaystyle 4444^{4444} \equiv X \equiv Y \equiv Z(mod9)$
From $\displaystyle Z \leq 12 $ and $\displaystyle Z \equiv 7(mod9)$ we have $\displaystyle Z=7$
$x = 4,444^{4,444} < 10,000^{4,444} \implies log_{10}(x) < log_{10}(10,000^{4,444}) \implies$

$log_{10}(x) < 4444 * 4 = 17,776.$

This does not imply that $x = 4,444^{4,444} < 180,000 \ \because \ 4,444^2 > 180,000.$

It does imply that $y \le 159,984 \implies z < 54.$

But I do not see how that tells you what z actually is.

Last edited by JeffM1; April 7th, 2018 at 11:49 AM.
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April 10th, 2018, 12:05 AM   #9
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Legendre's formula for the p-adic valuation of a number can be restated in terms of the sum of the digits of the number's p-adic expansion.

This would be a starting point I would use for an rigorous determination of an answer to a question regarding the sum of the digits of numbers.

https://en.wikipedia.org/wiki/Legendre%27s_formula
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