Ta có:
\(\overline{abcd}+\overline{abcd}=3576\)
\(\Rightarrow2\cdot\overline{abcd}=3576\)
\(\Rightarrow\overline{abcd}=\dfrac{3576}{2}=1788\)
\(\Rightarrow a=1;b=7;c=8;d=8\)
#\(Toru\)
=10 x abc + d + abc = 3576
= 11 x abc + d = 3576
= 3576 : 11 = abc ( dư d)
Ta có : 357 : 11 = 325 ( dư 1 )
abc= 325 , d = 1