Gọi số đó là \(\overline{ab}\)
\(\dfrac{ }{nabn}\)\(=\dfrac{ }{ab}\)x21
n000+\(\dfrac{ }{ab}\)x10+n=\(\dfrac{ }{ab}\)x21
n000+n=\(\dfrac{ }{ab}\)x11
nx1000+n=\(\dfrac{ }{ab}\)x11
nx(1000+1)=\(\dfrac{ }{ab}\)x11
nx1001=\(\dfrac{ }{ab}\)x11
nx1001:11=\(\dfrac{ }{ab}\)
nx91=\(\dfrac{ }{ab}\)
⇒n=1
⇒\(\dfrac{ }{ab}\)=91