This is just one method of several. Using approximations
color(brown)("Determine the starting point")
There are 5 digits in 19513
74xx10=740 not big enough
74xx100=7400 still not big enough
74xx1000=74000 5 digits but too big
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Try starting point
color(blue)("Step 1")
" "19513
100xx72->" "ul(7400) larr" don't like this. I can get closer!"
Notice that 2xx7=14 which is closer to the 19 part of 19513 so instead start from :
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color(blue)("Step 2")
" "color(white)(.)19513
200xx72->color(white)(..)ul(14800) larr" subtract"
" "4713
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color(blue)("Step 3")
6xx7=42 which is close to the 47 of 4713 so now we have:
" "color(white)(.)19513
200xx72->color(white)(..)ul(14800) larr" subtract"
" "4713
60xx72-> " "ul(4320) larr" subtract"
" "393
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color(blue)("Step 4")
5xx7=35 which is close to the 39 of 393 so we now have:
" "color(white)(.)19513
color(red)(200)xx72->color(white)(..)ul(14800) larr" subtract"
" "4713
color(red)(60)xx72-> " "ul(4320) larr" subtract"
" "393
color(red)(5)xx72->" "ul(360) larr" subtract"
" "33 larr" remainder" -> color(red)(33/72)
33 is less than 72 so we have finished. Unless you wish to go into decimal.
color(red)(200+60+5+33/72 = 265 33/72)
