First break down #589# into a sum of powers of #2#:
#589 = 512 + 64 + 8 + 4 + 1 = 2^9+2^6+2^3+2^2+2^0#
Next, repeatedly double #376509#, marking the multiples corresponding to the powers of #2# we found, then add them together:
#0color(white)(0000)color(blue)(376509)#
#color(white)(00000)753018#
#2color(white)(000)color(blue)(1506036)#
#3color(white)(000)color(blue)(3012072)#
#color(white)(0000)6024144#
#color(white)(000)12048288#
#6color(white)(00)color(blue)(24096576)#
#color(white)(000)48193152#
#color(white)(000)96386304#
#9color(white)(0)color(blue)(192772608)#
#color(white)(0000)color(blue)(376509)#
#color(white)(000)color(blue)(1506036)#
#color(white)(000)color(blue)(3012072)#
#color(white)(00)color(blue)(24096576)#
#color(white)(0)color(blue)(192772608)#
#"-----------------"#
#color(white)(0)color(blue)(221763801)#
This is #color(blue)(589 xx 376509)#
Multiply this sum by #80# by doubling it #3# times and adding a #0# at the end:
#color(white)(0)color(blue)(221763801)#
#color(white)(0)color(blue)(443527602)#
#color(white)(0)color(blue)(887055204)#
#color(blue)(1774110408)=>color(blue)(17741104080)#
Subtract #color(blue)(221763801)# to end up with #color(blue)(221763801 xx 79)#
#color(white)(-)color(blue)(17741104080)#
#-color(white)(000)color(blue)(221763801)#
#"---------------------"#
#color(white)(-)color(blue)(17519340279)#
