Let abc be a 3-digit number.
Find it, if acb + bca + bac + cba + cab = 3961.
We can write abc+acb+bca+bac+cba+cab= 3961+abc. The left side equals 222x(a+b+c). We need (a+b+c)>16, so 222x(a+b+c) is greater than 3961.
Trying by hand (a+b+c)=17, 18, 19, 20, we finally get 222x20=4440; 4440-3961=479; 4+7+9=20, so abc=479.