# TIFR2019-A-7

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What are the last two digits of $1! + 2! + \dots +100!$?

1. $00$
2. $13$
3. $30$
4. $33$
5. $73$

edited

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

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

selected

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.

edited by
1
2nd identity needs editing
0

@Satbir thanks for pointing .. edited now :)

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