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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.


Sagot :

[tex]N=1000w+100x+10y+z\\ w=z=1\\ x,y\in\{0,1,2,\ldots,7,8,9\}\\ 10x+y=\frac{N}{21}\\ 10x+y=\frac{1000w+100x+10y+z}{21}\\ 10x+y=\frac{1000+100x+10y+1}{21}\\ 10x+y=\frac{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} [/tex]