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I am a high school student and there is something I want to ask about the application of digital sums.

Let's say there is a fraction "520/7", let 520/7=a, so 520= a × 7, so if we now calculate the digital sums, it would be like 7= a × 7 , it means the digital sum of a should be 1 and nothing else so it means the remainder of this fraction on dividing by 9 is 1, but when we calculate the answer we see that it results in a repeating and infinite rational no. Which is 74.285714285714...and so on, which do not have any SINGLE digital sum as it keeps on changing as we add more and more digits. But our proof says it should be 1? So what's going on? Also when we say the digital sum of any no. is same as the remainder we get when we divide that no. By 9 but is it applicable for fractions as well? Because let's say there is a no. 18.225, if we divide this by 9 the remainder will be 0.225 and not 9, so this statement seems to be applicable on only integers. Am I right? I am not much aware about the modular arithmetic so please explain this in simpler terms. I just want to know can we apply the digital sum on fractions to verify if the result also have the same digital sum of not?

Shyam
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    There are several problems with your arguments. First, the "digital sum" is defined for natural numbers, not for fractions. Also, even when using natural numbers the digital sum of a product of two numbers is not the same as the product of the digital sum of the factors. In symbols, if we denote $\text{ds}(n)$ the digital sum of $n$, you are using that $\text{ds}(m\times n) = \text{ds}(m)\times \text{ds}(n)$, which is false in general. – jjagmath Mar 31 '24 at 13:28
  • But it is used several times like say 15 × 17, it's digital sum would be 6×8 which is 48 which finally gives 3, and if we check if it's answer really has digital sum 3, then it's answer is 255 and yes it has digital sum 3. So it is correct and is being derived for general no. On internet – Shyam Mar 31 '24 at 13:38
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    This is not clear. It's true that the iterated digit sum is multiplicative, but is that what your meant? The iterated digit sum essentially just returns the number $\pmod 9$ (instead of a remainder of $0$, though, you must use $9$). Beyond that, I'm not sure what you are asking. Lots of numbers of iterated digit sum $1$...any number of the form $9k+1$, to be precise. – lulu Mar 31 '24 at 13:44
  • Can we use digital sum for fractions? – Shyam Mar 31 '24 at 13:46
  • I mean there are questions which are very calculating in exams and then instead of solving the entire thing sometimes we just calculate their digital sum and then we eliminate all the answers from options which don't have same digital sum. So can we do this for fractions or only for integers this is what I mean? If we can use only for integers why it does work for fractions which are terminating? – Shyam Mar 31 '24 at 13:49
  • For non-integers, the three concepts mentioned here don't always match: 1) (iterated) (decimal) digital sum; 2) the remainder when divided by $9$, as in $\exists k\in \mathbb Z: \frac{520}7-\frac{16}7 = 9k$, but $\forall k\in\mathbb Z: \frac{520}7-1 \ne 9k$; 3) the modular fractions $\mod 9$ mentioned in the answer below. – peterwhy Mar 31 '24 at 18:17
  • @peterwhy Not true. They are all the same whenever the fractions are defined $!\bmod 9,,$ i.e. when writable with denominator coprime to $9,,$ as in my answer here. – Bill Dubuque Mar 31 '24 at 23:10
  • @BillDubuque I was referring to what the OP said, "if we divide this [18.225] by 9 the remainder will be 0.225 and not 9". Apparently by the OP's definition of remainder, a remainder of $0.225$ is different from a remainder of $0$. – peterwhy Mar 31 '24 at 23:20

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As explained here, modular fractions are well-defined if the denominator is coprime to the modulus, and they obey common fraction laws, e.g. $\!\bmod 9\!:\ 520\equiv 5\!+\!2\equiv 7\,\Rightarrow\, \frac{520}7\equiv \frac{7}7\equiv 1$. If you knew only the decimal we can still use this method to compute its value modulo $\,9\,$ by first first cancelling all factors of $3$ until we reach a denominator that's $\rm\color{#0a0}{coprime}$ to the modulus $9,\,$ i.e.

$\underbrace{74.\overline{\color{#0af}{285714}}_{\phantom{}}}_{\large 520/7} = \underbrace{74\!+\!\frac{\color{#0af}{285714}}{999999} = 74\!+\!\frac{10582}{\color{#0a0}{37037}}}_{\large {\rm cancel}\ 3^3}$ $\equiv 7\!+\!4\!+\!\frac{1+5+8+2}{3+7+3+7}\equiv 11\!+\!\frac{16}{20}^{\phantom{|^{|^|}}}\!\!\!\!\equiv\underbrace{ 2\!+\!\frac{\color{#c00}7}{2}\equiv 2\!-\!1}_{\large \color{#c00}7\ \equiv\ -2}$

In example $\,2\!:\ \,0.225 = \frac{225}{1000}\equiv \frac{2+2+5}{1+0+0+0}\equiv \frac{9}1\equiv \color{#0af}0,\,$ so $\,18.125\equiv 1\!+\!8\!+\!\color{#0af}0\equiv 0$.

Equivalently $\,18.125=\frac{729}{40}\equiv \frac{7+2+9}{4+0}\equiv \frac{18}{4}\equiv\frac{9}4\equiv 0\pmod{\!9}$.

Bill Dubuque
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  • For generalizations see this answer $\ \ $ – Bill Dubuque Mar 31 '24 at 15:12
  • I didn't understand it? Can u explain in a simple language like how does it solve the confusion? The answer doesn't actually have a single digital sum but the formula says it should( i am not a maths student, I just know basic maths like upto 10th grade, but the exam I will be giving requires to solve ques. Faster so I'll be using this digital sum to eliminate the options) – Shyam Mar 31 '24 at 15:21
  • @Shyam Are you not familiar with modular arithmetic (congruences)? – Bill Dubuque Mar 31 '24 at 15:25
  • Not fully, I just know very basic stuff like divisibility rules...here how and what does this proves? – Shyam Mar 31 '24 at 15:39