#color(blue)("Method 1 - a traditional approach")#

Really this is long division but perhaps in a different format to what you are used to seeing.

Starting point: Note that #5xx8=40# and 4 is the first digit on the left.

#" "4009#

#color(magenta)(800)(5) ->ul(4000)larr" subtract"#

#" "9#

#color(magenta)(color(white)(..)1)(5)->color(white)(000)ul(5) larr" subtract"#

#" "4#

#color(magenta)(color(white)(.)4/5)(5)->" "ul(4) larr" subtract"#

#" "0 larr" as this is 0 we stop. No remainder"#

So the answer is: #color(magenta)(800+1+4/5 = 801 4/5#

But #4/5->8/10=0.8" "# giving: #" "color(magenta)(801.8)#

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#color(blue)("Method 2 -A sort of non calculator cheat method")#

#-:5# is the same as #xx1/5#

But #1/5# is the same as #" "2xx1/10#

So write #4009xx1/5" "# as #" "4009xx2xx1/10#

#=8018xx1/10=801.8#