Question

Both the leftmost digit and the rightmost digit of a four-digit number

N are equal to 1. When these digits are removed, the two-digit number
thus obtained is N ÷ 21. Find N.

Answer 1

`N=1000w+100x+10y+z\\ w=z=1\\ x,y\in\{0,1,2,\ldots,7,8,9\}\\ 10x+y=(N)/(21)\\ 10x+y=(1000w+100x+10y+z)/(21)\\ 10x+y=(1000+100x+10y+1)/(21)\\ 10x+y=(100x+10y+1001)/(21)\\ 210x+21y=100x+10y+1001\\ 110x+11y-1001=0\\ 10x+y-91=0\\ y=-10x+91\\\\ \hbox{The above equation meets the condition }x,y\in\{0,1,2,\ldots,7,8,9\}\\ \hbox{only for } x=9:\\ y=-10\cdot9+91\\ y=-90+91\\ y=1\\\\ \hbox{Therefore:}\\ N=1000\cdot1+100\cdot9+10\cdot1+1\\ N=1000+900+10+1\\ N=\boxed{1911}`