The ten's digit in 1 !+4 !+7 !+10 !+12 !+13 !+15 !+16 !+17 ! is divisible by

Step-by-step solution

We have, aligned &1 !=1,4 !=24,7 !=5040,10 !=3628800 , &11 !=39916800 aligned So, last two digit from 10 t is always 00 . Now, 1 !+4 !+7 !=1+24+5040=5065 So, ten's digit in given addition is 6 which is divisible by 31 .