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)#