9 votes

Best answer

$1!=1, 2!=2, 3!=6, 4!=24, 5!=120.$

After $4,$ factorials of all numbers end in $0.$

So, **unit digit = Unit digit of (1+2+6+4) = Unit digit of 13 = 3**.

$6!=720, 7!= \_\;\_40, 8!=\_\;\_\;20, 9!=\_\;\_\;80, 10! = \_\;\_\;\_\;800.$

After $9,$ factorials of all numbers end in $00.$

**Tens digit = Unit digit of (1+2+2+2+4+2+8) = Unit digit of 21 = 1**.

**Last 2 digits = 13**.

Source: https://www.quora.com/What-are-the-last-2-digits-of-1-2-3-100

4 votes

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.