a)4a1b/2 si a+b=5
4114/2 1+4=5
4312/2 3+2=5
b)21ab/2 si a=3b b<a
2162/2 a=2x3
c)1ab/5 a=b+3 b(0,5)
130 0+3=3
185 5+3=8
d)3a2b/10 pentri a fi divizibil cu 10 b trebuie sa fie 0
a=2x0+1=1
3120/10
E)abcd/10 si dcba numere pare consecutive
dcba=0246
abcd=6420/10
f)abcd/5
dcba=consecutive
dcba=0123
dcba=5678
abcd/5=3210
abcd/5=8765