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Aratati ca numarul A=1×2×3×....×2017+2017+2017 la puterea 2+....+2017 la puterea 2017 nu este patrat perfect.
Ma poate ajuta cineva ?


Răspuns :

2017 1 se termija in 0
U(2017+2017²+2017³+2017^4)=U(7+9+3+1)=8
U (2017+...+2017^2016)=U(504*8) =2
u (2017^2017)=U(2017^1)=7

U (0+8+2+7)=U(17)=7
dar nici un p.p.nu setermina in 7
deci A nu e p.p.