Question

Admitere automatica Gheorghe Asachi Iasi 2018

Answer 1

 

U(1!) = 1

U(2!) = 2

U(3!) = U(1×2×3) = 6

U(4!) = U(1×2×3×4) = U(24) =4

U(5!) = U(1×2×3×4×5) = U(120) = 0

Incepand de la 5! pana la 2018! toate au ultima cifra = 0.

⇒ U(1! + 2! + 3! + ... + 2018!) =

= U(1 + 2 + 6 + 4 + 0 + 0 + ... + 0) = U(13) = 3