April 6th, 2018, 01:09 PM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 227 Thanks: 26  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. 
April 6th, 2018, 03:46 PM  #2 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,619 Thanks: 845 
Ever heard of google? https://www.google.ca/search?source=....0.KCxqWg0Hexk 
April 6th, 2018, 05:02 PM  #3 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,797 Thanks: 715 Math Focus: Wibbly wobbly timeywimey stuff.  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, 05:15 PM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,619 Thanks: 845 
Suggested google as a way to learn...

April 6th, 2018, 06:08 PM  #5  
Senior Member Joined: Sep 2015 From: USA Posts: 1,980 Thanks: 1027  Quote:
I don't believe OP is even a student. I'm not sure what their purpose here is.  
April 6th, 2018, 08:23 PM  #6  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,797 Thanks: 715 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
April 7th, 2018, 04:13 AM  #7 
Senior Member Joined: Dec 2015 From: Earth Posts: 227 Thanks: 26 
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, 11:22 AM  #8  
Senior Member Joined: May 2016 From: USA Posts: 1,038 Thanks: 423  Quote:
$log_{10}(x) < 4444 * 4 = 17,776.$ 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.  
April 10th, 2018, 12:05 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 padic valuation of a number can be restated in terms of the sum of the digits of the number's padic 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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