To find the last $2$ digits of $1!+2!+3!....+100!$
we shall find $(1!+2!+3!....+100!)mod 100$
I will be using the following $2$ identities repeatedly..
$(a+b)mod m=(amod m +$b$mod m) mod$ $m$
$(a*b)mod m=(amod m * b mod m)mod m$
$(1!+2!+3!....+100!)mod 100$
$=$$(1! mod100 + 2!mod 100 + 3! mod 100+....+100!mod100)mod 100$
$1! mod 100=1$
$2! mod 100=2$
$3! mod 100=6$
$4! mod 100=24$
$5! mod 100=(4!mod 100 *5 mod 100)mod 100=(24*5)mod 100=120 mod 100=20$
$6! mod 100=(6*5!)mod100=(6 mod 100*5! mod100)mod100=(6*20)mod100=20$
$7! mod 100=(7*6!)mod100=(7 mod 100*6! mod100)mod100=(7*20)mod100=40$
$8! mod 100=(8*7!)mod100=(8 mod 100*7! mod100)mod100=(8*40)mod100=20$
$9! mod 100=(9*8!)mod100=(9 mod 100*8! mod100)mod100=(9*20)mod100=80$
$10! mod 100=(10*9!)mod100=(10 mod 100*9! mod100)mod100=(10*80)mod100=0$
All factorials after $10!$ can be expressed as $(k*10!)$ where $k$ is a integer
for example,$13! =(13*12*11*10!)$ here $k=13*12*11$
$16!=(16*15*14*13*12*11*10!)$ here $k=16*15*14*13*12*11$
$n!$ mod $100$ (where $n>10$)$=(k*10!)mod 100=(k mod 100 * 10! mod 100)=(k mod 100 * 0)=0$
So all factorials of all integers $>10\ mod\ 100$ are $0$
.i.e., $11!\ mod\ 100 =12!\ mod\ 100 = 13!\ mod\ 100.....=100!\ mod\ 100=0$
$\therefore$ $(1!+2!+3!....+100!)mod 100$
$=$$(1+2+6+24+20+20+40+20+80+0+0+0.........+0)mod 100$
$=213\ mod\ 100$
$=13$
Hence Option $B.$ is correct choice.
references:https://brilliant.org/wiki/modular-arithmetic/