+20 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
+26367 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:
