abc3 - 3abc = 3581

=> abc x 10 + 3 - 3000 + abc = 3581

=> abc x (10+1) + 3 - 3000 = 3581

=> abc x (10+1) +3 = 3581 + 3000

=> 11 x abc + 3 = 6581

=> 11 x abc = 6581 - 3

=> 11 x abc = 6578

=> abc = 6578 : 11

=> abc = 598

Vậy abc = 598

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Lời giải:
$\overline{abc3}+1620=\overline{3abc}$

$\overline{abc}\times 10+3+1620=3000+\overline{abc}$
$\overline{abc}\times 10+1623=3000+\overline{abc}$

$\overline{abc}\times 10-\overline{abc}=3000-1623$

$9\times \overline{abc}=1377$

$\overline{abc}=1377:9=153$ 

Vậy số cần tìm là $153$

\(\overline{abc3}-\overline{3abc}=387\)

\(\Rightarrow10.\overline{abc}+3-3000-\overline{abc}=387\)

\(\Rightarrow9.\overline{abc}=3384\Rightarrow\overline{abc}=3384:9=376\)

abc=153

abc=780

Ta có: 

\(\overline{abc3}=\overline{abc}\times10+3\)

\(\overline{6abc}=6000+\overline{abc}\)

Phép tính trở thành:

\(\overline{abc}\times10+3+4125=6000+\overline{abc}\)

\(\overline{abc}\times10+4128=600+\overline{abc}\)

\(\overline{abc}\times10-\overline{abc}=6000-4128\)

\(\overline{abc}\times\left(10-1\right)=1872\)

\(\overline{abc}\times9=1872\)

\(\overline{abc}=1872:9=208\)

Vậy: \(a=2,b=0,c=8\)