Ta có: abcd+a+b+c+d=2013
=>a.1000+b.100+c.10+d+a+b+c+d=2013
=>a.1001+b.101+c.11+d.2=2013
=>a.1001<2013
=>a<3
mà 0<a<3
=>a=1,2
*Xét a=1=>1001+b.101+c.11+d.2=2013
=>b.101+c.11+d.2=1012
=>b.101=1012-(c.11+d.2)
Vì c<10, d<10
=>c.11<110, d.2<20
=>c.11+d.2<130
=>1012-(c.11+d.2)>1012-130
=>b.101>882
=>b>8
mà 8<b<10=>b=9
=>909+c.11+d.2=1012
=>c.11+d.2=103
=>c.11=103-d.2
mà d.2<20
=>103-d.2>103-20=83
=>c.11>83
=>c>7
mà 7<c<10=>c=8,9
=>103 lẻ, d.2 chẵn=>103-d.2 lẻ
=>c lẻ=>c=9
=>d=103-9.11=103-99=4
=>abcd=1994
Thử lại: abcd+a+b+c+d=1994+1+9+9+4=2017 khác 2013
=>loại
*Xét a=2=>2002+b.101+c.11+d.2=2013
=>b.101+c.11+d.2=11
=>b.101<11
=>b<1
=>b=0
=>0+c.11+d=11
=>c.11+d.2=11
=>c.11=11-d.2
=>c.11<_11=>c<_1
=>c=0,1
mà 11 lẻ, d.2 chẵn=>11-d.2 lẻ
=>c.11 lẻ=>c lẻ
=>c=1=>11+d.2=11=>d.2=0=>d=0
Vậy abcd=2010
