Question

efectuati:
11+12+13+...+95
5+8+11+...+305
stabileste paritatea numerelor :
S1  2+4+6+...+198
S2 1+3+5+...+175
S3 5+9+13+...+101
S3 5+10+15+...+85
determinati ultima cifra al produsului:
P1 1×3×5×...×71
P2 1×2×3×4×...×35
P3 1357×8642
Va rog ajutati-ma ! 
Repede!!!

Answer 1

11+12+13+...+95
(1+2+3+...+95)-(1+2+3+...+10)=
4560-55=4505
5+8+11+...+305
5+(5+1×3)+(5+2×3)+(5+3×3)+...+(5+100×3)=
101×5+5×100×101:2=
505+25250=25755
S1=  2+4+6+...+198
2×(1+2+3+...+99)=9900
S2=  1+3+5+...+175
(1+2+3+...+175)-(2+4+6+...+174)=
15400-2×(1+2+3+...+87)
15400-7656=7744
S3=  5+9+13+...101
5+(5+1×4)+(5+2×4)+(5+3×4)+...+(5+24×4)=
25×5+4×(1+2+3+...+6)=
120+84=204
S4 = 5+10+15+...+85
5×(1+2+3+...+17)=765

Answer 2

S= 11+12+13+.......+95   notam  S1= 1+2+3+......95 = 95·96/2=4560          si S2 = 1+2+3+......+10 =10·11/2 =55 ⇒  S = S1 - S2 = 4505

5=3·0 + 5

8= 3·1+5

11=3·2+5

...............

305 =3·100+5 ⇒S = 3(1+2+3+.....+100) + 5·101 = 3·100·101/2 +505 = 15158 +505 = 15663

S1 = 2(1+2+3+.....+99) = 2· 99·100 /2 = 9900

S2 = 1+3+5+.......+175 = 176·88 = 15488    (176= 1+175 =3+173=...)

5 = 4·0 +5

9 =4·1+5

13=4·2+5

,,,,,,,,,,,,,,,,

101=4·24+5 ⇒ S3 = 4(1+2+3+.....+24)+ 5·25 = 4·24·25 /2 +125=1325

S4 = 4(1+2+3+.......+17) = 4·17·18 /2 =612

Uc(P1) = Uc[(1·3·5·7·9)×(11·13·15·17·19)×......(61·63·65·67.69)·71] = Uc(5·5·5·.....·1) =5

Uc(P2) = Uc[(1×2×3×4×5×6×7×8×9×10)·      ·(31·32·33·34·35) = 0

Uc(P3) = 4