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It should work like this:
\(f(5, 3) = 555\)

\(f(12,3)=121212\)

 

\(555=5\cdot10^2+5\cdot10^1+5\cdot10^0\)

\(121212=12\cdot10^4+12\cdot10^2+12\cdot10^0\)

 

We see that the power of ten dependence on the length of the number that we're repeating.

Knowing this we can bring out the formula.

\(rep\left(x,n\right)=\sum_{c=0}^{n-1}x\cdot10^{c\lfloor\log10x\rfloor}\)

Where \(x\in\mathbb{N}\) (x is the repeating number)

Where \(n\in\mathbb{N}\) (n is the number of repetitions)

\(\lfloor\operatorname{log}10x\rfloor\) - length of x

 

\(rep\left(5,3\right)=555\)

\(rep\left(12,3\right)=121212\)

 

Look at this on Desmos: https://www.desmos.com/calculator/4baj3yaxaz

off-topic
 Sep 23, 2019
edited by JoshuaGreen  Oct 22, 2019
 #1
avatar+23850 
+3

It should work like this:
\(f(5, 3) = 555 \\ f(12,3)=121212 \\ \\ 555=5\cdot10^2+5\cdot10^1+5\cdot10^0 \\ 121212=12\cdot10^4+12\cdot10^2+12\cdot10^0 \)

 

We see that the power of ten dependence on the length of the number that we're repeating.

Knowing this we can bring out the formula.

\(rep\left(x,n\right)=\sum_{c=0}^{n-1}x\cdot10^{c\lfloor\log_{10}x\rfloor}\)

Where  (x is the repeating number)

Where  (n is the number of repetitions)

\(\lfloor log_{10}x\rfloor\) - length of x

 

Sorry there is a mistake:

The length of \(x\) is \(\mathbf{\lfloor log_{10}x\rfloor +\color{red} 1 }\)

\(\large{rep\left(x,n\right)=\sum \limits_{c=0}^{n-1}x\cdot10^{c \left( \lfloor\log_{10}x\rfloor+1 \right) }}\)

see Desmos:

 

hint:

\(\begin{array}{|rcll|} \hline &&\mathbf{ \lfloor\log_{10}x\rfloor+1 } \\ &=& \lfloor\log_{10}x+1 \rfloor \\ &=& \lfloor\log_{10}x+\log_{10}(10) \rfloor \\ &=&\mathbf{ \lfloor\log_{10}(10x) \rfloor } \\ \hline \end{array}\)

see Desmos:

 

laugh

 Sep 24, 2019
 #2
avatar
+1

 Thank you again. Your responses are really great. Every time you give me such useful information. Where do you get these hints? I need to know them all!

Guest Sep 24, 2019
 #3
avatar+20 
+1

The guest above is me - JoshuaGreen.  Forgot log in.

JoshuaGreen  Sep 24, 2019
 #4
avatar+23850 
+2

Thank you, JoshuaGreen !

 

laugh

heureka  Sep 25, 2019

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