Question

Va rog urgent determinati ultima cifra a nr 8^1223, 9^203, 3•2^35+2•5^100. Aratati ca a este patrat perfect a=(1+2+3+...99)x2+100
a=1+2+3+...+50-25x50
a=3+6+9+...+75-75
Aflati x:(3x)^5=9^5x3^2x3^3
Va rog

Answer 1

U(8^1223)=8
U(9^203)=9
3×2^25+2×5^100=Nu stiu

a=(99×100:2)×2+100
a=(99×50)×2+100
a=4.950×2+100
a=9.900+100
a=10.000

a=50×51:2-25×50
a=25×51-25×50
a=25×(51-50-1)
a=25×0=0

a=1+2+3+...+25-75
a=25×26:2-75
a=25×13-75
a=975-75=900

x:(3x)^5=(3^2)^5×3^2×3^3
x:(3x)^5=3^10×3^2×3^3
x:(3x)^5=3^15
Scz de aici nu mai ştiu

Sper că te-am ajutat