The ten's digit in $1 !+4 !+7 !+10 !+12 !+13 !+15 !+16 !+17 !$ is divisible by
The ten's digit in $1 !+4 !+7 !+10 !+12 !+13 !+15 !+16 !+17 !$ is divisible by
4 !
3 !
5 !
7 !
Solution
$\begin{gathered}
\text { } 1 !+4 !+7 !=1+24+5040 \\
1 !+4 !+7 !=5065 \\
(10 !+12 !+\ldots+17 !) \text { this value }
\end{gathered}$
have last 2 digits as zero es
So, $10^{\prime}$ 's digit of given question is same as $10^{\prime} \mathrm{s}$ digit of $(1 !+4 !+7 !)$
$\therefore$ Required 10 's digit is $6=3$ !
Hence, option (1) is correct.