Using approximation with trial and error.
Notice that #3xx7=21# implying that the most significant figure could be 3
Consider the #color(red)(1)" from 7"color(red)(1)#. So 1 times something should give us the last digit in #color(brown)(2165)color(green)(5)# and that something is 5. So the last digit in what we are multiplying by is 5.
#color(blue)("Try "35xx71)->" "71 #
#" "underline(color(white)(..)35) larr"multiply"#
#" "2130 larr "71 multiplied by 30"#
#" "underline(color(white)(..) 355) larr"71 multiplied by 5"#
#" "2485" "larr 2130+355#
#color(blue)(2485 " is too small so does not work")#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Try "305xx71)->" "305 #
#" "underline(color(white)(..)71) larr" multiply"#
#" "21350 larr"305 multiplied by 70"#
#" "underline(color(white)(..)305) larr"305 multiplied by 1"#
#" "21655 larr 21350+305#
#color(blue)(21655 " is exactly correct")#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So #305xx71 =21655#
So #21655-:71=305#
