c)[9(1+2+3+....+100)-7(1+2+3+.....+100)]:(1+2+3.....+100)=
(1+2+3+....+100)(9-7):(1+2+3+....+100)=
se simplifica (1+2+3+.....+100)
rezultatul fiind (9-7)=2
2)2^77+2^76+2^75+2^74 divizibil cu 15
2^74(2^3+2^2+2^1+1)
2^74(8+4+2+1)
15|2^74×15
3)a)a=7+7-21:3
a=14-7
a=7
b=9-3[9-3(3-3:3)]
b=9-3(9-3×2)
b=9-3×3
b=0
c=[6^0×8^2-2^5):2^3+1]×2-9
c=[(1×64-32):8+1]×2-9
c=(32:8+1)×2-9
c=5×2-9
c=1
b)a divizibil cu 5(F)
c)c^2013+b^a=1^2013+0^7=1+0=1
