Question Number 48884 by tanmay.chaudhury50@gmail.com last updated on 29/Nov/18
$${find}\:{last}\:{digit}\:{of} \\ $$$$\mathrm{1}!+\mathrm{2}!+\mathrm{3}!+\mathrm{4}!+...+\mathrm{23}! \\ $$$${and}\:\mathrm{1}!+\mathrm{2}!+\mathrm{3}!+...+\mathrm{134}! \\ $$
Answered by mr W last updated on 29/Nov/18
$${last}\:{digit}\:{of}\:{n}!\:{is}\:\mathrm{0}\:{for}\:{n}\geqslant\mathrm{5}. \\ $$$$\mathrm{1}!+\mathrm{2}!+\mathrm{3}!+\mathrm{4}!=\mathrm{33} \\ $$$$\Rightarrow{last}\:{digit}\:{of}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}!\:{is}\:{always}\:\mathrm{3}\:{with}\:{n}\geqslant\mathrm{5}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 29/Nov/18
$${thank}\:{you}\:{sir}...\:{i}\:{have}\:{posted}\:{this}\:{question}\:{as} \\ $$$${it}\:{is}\:{a}\:{tricky}\:{question}... \\ $$
Commented by mr W last updated on 29/Nov/18
$${nice}\:{question}! \\ $$