We need to solve #1710 -: 6#.
My first step would be to either use long division or expanding to simplify. I'll do both
First I'll expand the numbers. The basis of this is to find the largest number that can divide with the #6# (or the factors of #2# or #3# inside it)
#1710/6#=#(cancel(2)*855)/(cancel(2)*3)#=#855/3#=#(cancel(3)*285)/cancel3#=#285#.
For this system, we just keep breaking down #1710# until the divisor just becomes #1#. This works for "nicer" numbers best, and for such a strange number as #1710# you'd probably want a calculator.
Long division requires nothing but pencil and paper:
#color(white)(6)color(white)(|)color(white)(-)color(black)(0)color(black)(2)color(black)(8)color(black)(5)#
#color(white)(6)color(black)(---)#
#6|color(white)(-)color(black)(1)color(black)(7)color(black)(1)color(black)(0)#
#color(white)(6)color(white)(|)color(black)(-)color(black)(1)color(black)(2)color(white)(1)color(white)(0)#
#color(white)(6)color(white)(-)color(black)(--)#
#color(white)(6)color(white)(|)color(white)(-)color(white)(1)color(black)(5)color(black)(1)color(white)(0)#
#color(white)(6)color(white)(|)color(white)(1)color(black)(-)color(black)(4)color(black)(8)color(white)(0)#
#color(white)(6)color(white)(--)color(black)(--)#
#color(white)(6)color(white)(|)color(white)(-)color(white)(1)color(white)(4)color(black)(3)color(black)(0)#
#color(white)(6)color(white)(|)color(white)(1)color(white)(4)color(black)(-)color(black)(3)color(black)(0)#
#color(white)(6)color(white)(-.-)color(black)(--)#
#color(white)(6)color(white)(|)color(white)(1)color(white)(4)color(white)(-)color(black)(0)color(black)(0)#
We get the same answer. So, either way works.
