If you have the expression 6÷8=27 how can you insert the same number into the equation twice in order to make the equation correct (no addition, multiplication, division or subtraction, just the idea that having 6 and 2 would make 62)
I've tried putting some numbers after the 6 that are below 6 so that when I put the same number before the 8 the answer will be greater that 1. i.e. 628÷288 but obviously that doesn't work. I cannot seem to come up with a method to do this sort of problem generally either
If we insert the same number $X$ after $6$ and before $8$ we have the equation $6X\div X8=27$, that is: $$ 6\cdot10^k+X=27(10X+8), \quad\hbox{or}\quad 269X=6\cdot10^k-216, $$ where $k$ is the number of digits of $X$. This has the smallest integer solution for $k=128$: $$ \begin{align} X=&02230483271375464684014869888475836431226765799256505576208178438\\ &661710037174721189591078066914498141263940520446096654275092936, \end{align} $$ where I inserted a leading $0$ to have a total of $128$ digits.