a+b=47
ba+dc=173
deci
10a+b+10c+d=47 => 10(a+c)+(b+d)=47 => b+d=47- 10(a+c)
10b+a+10d+c=173 => (a+c)+10(b+d)=173
=> a+c+10[47-10(a+c)]=173
a+c+470-100a-100c=173
297=99a+99c
a+c=3, dar a,d≠0, deci a=1 si c=2
b+d=47-30=17 , deci b=8 si d=9
Verificare:
18+29=47
81+92=173!
