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 June 28th, 2019, 12:53 PM #1 Newbie   Joined: Jun 2019 From: Italy Posts: 7 Thanks: 0 Sum of number's number Hey, I have a number theory problem: determine the sum of the digits of a natural number. I have been thinking about it for a lot time during this days, but I can't still find a solution. Can someone give me an idea or advise me a book/internet site with something useful? Thank a lot. June 28th, 2019, 01:48 PM #2 Global Moderator   Joined: May 2007 Posts: 6,823 Thanks: 723 Add them up? What are you looking for? June 28th, 2019, 03:33 PM #3 Senior Member   Joined: Jun 2019 From: USA Posts: 223 Thanks: 99 I can find it modulo 9 no problem.  June 28th, 2019, 05:52 PM #4 Member   Joined: Oct 2018 From: USA Posts: 93 Thanks: 66 Math Focus: Algebraic Geometry This seems pretty brute force, wouldn't surprise me at all if there was something more elegant. Let each digit $d_i \in \{0,1,2 \dots , 9\}$ be such that $\displaystyle d_{1}d_{2}d_{3} \dots d_{N} = a \in \mathbb{N}$ (This is concatenation, but also $\displaystyle a= \sum_{i=1}^{N} d_{i}10^{(N-i)}$ ) Then we know $d_1 = floor \left(10^{-(N-1)}a \right)$ Now, we know $a-10^{(N-1)}d_{1} = d_{2}d_{3} \dots d_{N}$ thus $d_{2} = floor \left(10^{-(N-2)} \left(a-10^{(N-1)}d_{1} \right) \right)$ and $d_{3} = floor \left(10^{-(N-3)} \left(a-10^{(N-1)}d_{1} - 10^{(N-2)}d_{2} \right) \right)$ Continuing this strategy leaves $\displaystyle d_{n} = floor \left( 10^{-(N-n)} \left( a - \sum_{i=1}^{n-1} d_{i}10^{(N-i)} \right) \right)$ So, the sum of digits should be $\displaystyle \sum_{i=1}^{N}d_{i} = floor \left(10^{-(N-1)}a \right) + \sum_{n=2}^{N} floor \left( 10^{-(N-n)} \left( a - \sum_{i=1}^{n-1} d_{i}10^{(N-i)} \right) \right)$ For $N \geq 2$ Thanks from topsquark and Vitty Last edited by Greens; June 28th, 2019 at 06:06 PM. Reason: Negative Signs June 28th, 2019, 11:34 PM   #5
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Thanks for your answer. I tried with another different way:

-Let $\displaystyle d_i \in \{0,1,2 \dots , 9\}$ and $\displaystyle \displaystyle d_{1}d_{2}d_{3} \dots d_{N} = a \in \mathbb{N}$

-Now I know that each digit is generated by: $\displaystyle d_i = \sum_{i=1}^{\left \lfloor log_{10}(A)+1 \right \rfloor}\frac{A\,\, mod\,\, 10^{i}-A\,\, mod\,\, 10^{i-1}}{10^{i-1}}$

-I expand the sum and I obtain: $\displaystyle \frac{A\, mod\,\, 10-A\,\, mod\,\, 1}{1}+\frac{A\, mod\,\, 10^{2}-A\,\, mod\,\, 10}{10}+\cdots +\frac{A\, mod\,\, 10^{i}-A\,\, mod\,\, 10^{i-1}}{10^{i-1}}$

-I have to simplify the denominator so: $\displaystyle \frac{1\cdot (A\, mod\,\, 10-A\,\, mod\,\, 1)}{1}+\frac{10\cdot (\frac{A}{10}\, mod\,\, 10-\frac{A}{10}\,\, mod\,\, 1)}{10}+\cdots +\frac{10^{i-1}\cdot (\frac{A}{10^{i-1}}\, mod\,\, 10-\frac{A}{10^{i-1}}\,\, mod\,\, 1)}{10^{i-1}}$

-I split the term: $\displaystyle A\, \, mod\, \, 10+\frac{A}{10}\, \, mod\, \, 10+\cdots+\frac{A}{10^{i-1}}\, \, mod\, \, 10-(A\, \, mod\, \, 1+\frac{A}{10}\, \, mod\; 1+\cdots +\frac{A}{10^{i-1}}\, \, mod\, \, 1)$

-Now I would have picked up $\displaystyle mod\,\,10$ and $\displaystyle mod\,\,1$: $\displaystyle \left (\sum_{i=1}^{\left \lfloor log_{10}(A)+1 \right \rfloor}\frac{A}{10^{i-1}}\right)\, \, mod\, \, 10-\left (\sum_{i=1}^{\left \lfloor log_{10}(A)+1 \right \rfloor}\frac{A}{10^{i-1}}\right)\, \, mod\, \, 1$; but I can't; some ideas to move forward?

Quote:
 Originally Posted by Greens $\displaystyle \sum_{i=1}^{N}d_{i} = floor \left(10^{-(N-1)}a \right) + \sum_{n=2}^{N} floor \left( 10^{-(N-n)} \left( a - \sum_{i=1}^{n-1} d_{i}10^{(N-i)} \right) \right)$
Can we move from this sum to a function without $\displaystyle \sum$? Tags function, number, number theory, sum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Germann Advanced Statistics 1 March 12th, 2019 09:34 AM thinhnghiem Math 0 May 15th, 2018 08:07 AM thanhvinh0906 Advanced Statistics 3 August 30th, 2017 04:27 PM Shen Elementary Math 2 June 5th, 2014 07:50 AM UnreasonableSin Number Theory 2 June 13th, 2010 12:03 AM

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