Fie n=abcd, numar de 4 cifre; abcd=1000a+100b+10c+d
suma cifrelor= s(n)= a+b+c+d
n-s(n)=1000a+100b+10c+d-(a+b+c+d)
999a+99b+9c=2007
9(111a+11b+c)=9•223
111a+11b+c=223
=> a=2 ; b=0; c=1; d∈{0,1,2,3,4....9}
abcd ∈{2010, 2011, 2012, 2013, ......, 2019}
n∈{2010, 2011, 2012, 2013, ......, 2019}