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 April 6th, 2018, 02:09 PM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 256 Thanks: 28 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 02:12 PM.
 April 6th, 2018, 04:46 PM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,448 Thanks: 946 Ever heard of google? https://www.google.ca/search?source=....0.KCxqWg0Hexk
April 6th, 2018, 06:02 PM   #3
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Quote:
 Originally Posted by Denis 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

 April 6th, 2018, 06:15 PM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,448 Thanks: 946 Suggested google as a way to learn... Thanks from topsquark
April 6th, 2018, 07:08 PM   #5
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Quote:
 Originally Posted by topsquark 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.

April 6th, 2018, 09:23 PM   #6
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 Originally Posted by romsek 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

 April 7th, 2018, 05:13 AM #7 Senior Member   Joined: Dec 2015 From: Earth Posts: 256 Thanks: 28 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 $\displaystyle Z=7$
April 7th, 2018, 12:22 PM   #8
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Quote:
 Originally Posted by idontknow 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 12:49 PM.

 April 10th, 2018, 01:05 AM #9 Banned Camp   Joined: Apr 2016 From: Australia Posts: 244 Thanks: 29 Math Focus: horses,cash me outside how bow dah, trash doves 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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