Writing numbers in base 10 means to write them as a combination of powers of 10, for example[tex]1925 = 1\cdot 10^3 + 9\cdot 10^2 + 2\cdot 10^1+5\cdot 10^0[/tex]So, writing numbers in any base means to do the same thing: for example, in base 3, we have[tex]102_3 = 1\cdot 3^2 +0\cdot 3^1 +2\cdot 3^0 = 12_{10}[/tex]So, in order to write 80 in base 3, we have to find out the largest power of 3 that fits: we have[tex]3^0=3,\ \ 3^1=3,\ \ 3^2=9,\ \ 3^3=27,\ \ 3^4=81[/tex]So, since [tex]3^4[/tex] is too much, our number will have 4 digits:[tex]abcd_3 = a\cdot 3^3 +b\cdot 3^2 +c\cdot 3^1 + d\cdot 3^0[/tex]In order to find the digits, again we see "how much it fits": we have[tex]0\cdot 3^3 = 0\ \ 1\cdot 3^3 = 27,\ \ 2\cdot 3^3 = 54[/tex]So, we choose [tex]a=2[/tex]. The remainder is [tex]80-54=26[/tex]Now we need to fix the coefficient for [tex]3^2[/tex]: we have[tex]0\cdot 3^2 = 0\ \ 1\cdot 3^2 = 9,\ \ 2\cdot 3^2 = 18[/tex]Again we choose [tex]b=2[/tex]Keep going like this and you'll find out that[tex]80_3 = 2222[/tex]This was actually a special case, because our number is exactly one less than a power of 3: we have[tex]80=3^4-1[/tex]and just like in base 10, when we subtract 1 from a power of 10 we get a number composed by 9 only:[tex]10000000-1 = 9999999[/tex]In every base, when we subtract 1 from a power of the base we get a number composed by (base-1) only:[tex] (10000-1)_3 = 2222[/tex]