Menu Close

Question-171342




Question Number 171342 by mathlove last updated on 13/Jun/22
Commented by infinityaction last updated on 13/Jun/22
3
$$\mathrm{3} \\ $$
Answered by Rasheed.Sindhi last updated on 13/Jun/22
A=1!+2!+3!+4!+5!+6!...+1000!_(duisible by 5)   A≡1!+2!+3!+4!=33(mod 5)  A≡3(mod 5)
$${A}=\mathrm{1}!+\mathrm{2}!+\mathrm{3}!+\mathrm{4}!+\underset{{duisible}\:{by}\:\mathrm{5}} {\underbrace{\mathrm{5}!+\mathrm{6}!…+\mathrm{1000}!}} \\ $$$${A}\equiv\mathrm{1}!+\mathrm{2}!+\mathrm{3}!+\mathrm{4}!=\mathrm{33}\left({mod}\:\mathrm{5}\right) \\ $$$${A}\equiv\mathrm{3}\left({mod}\:\mathrm{5}\right) \\ $$
Commented by mathlove last updated on 13/Jun/22
thanks
$${thanks} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *