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
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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 :
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 2")#
#" "color(white)(.)19513#
#200xx72->color(white)(..)ul(14800) larr" subtract"#
#" "4713#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#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#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#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)#
