Sum! Divided by 100

• We know that 1! is 1, 2! is 2, 3! is 3, 4! is 24 and 5! is 120 by very simple mental arithmetic. It is obvious that the numbers are getting larger very quickly.

• We can also see very easily that if k! is divisible by 100 then (k+1)! must be divisible by 100 as they will share the set of factors that multiply to give 100.

• Therefore as soon as we find the smallest number, n! that is perfectly divisible by 100 we know that all the numbers that follow must be also divisible by 100.

• We can do this very quickly through long multiplication as it is only the last two digits we need to know to see if a number is divisible by 100. (they are both zeros)

• This number is 10! (doesn't take very long to find out, does it?) We can then find the sum of the last two digits of all the previous numbers starting with 9! (taking the two together as a number in their own right) and divide this sum by 100.

• This is 1+2+6+24+20+20+40+(20+80) we can ignore the last two as they very obviously sum together to make 100. We get an answer of 113 - so the answer is 13.

When a number is divided by 100 the remainder is the last 2 digits. For $n\geq 10$ , $n!$ and above it will end with two zeros [Why? -Calvin]. So when you divide such numbers we get remainder as zero. So for us the remainder will only come from 1! to 9!. Calculating, 1! gives remainder 1,2! gives remainder 2,3! gives 6, 4! gives 24,5! gives 20,6! gives 20, 7! gives 40, 8! gives 20 and 9! gives 80. Add up all numbers we get $1+2+6+24+20+20+40+20+80=213$ . Hence 13 is the remainder.

Since 10! ends in 00, we only have to look at the remainder of when 1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! is divided by 100 to get the remainder when whole series is divided by 100 since every term past 9! is always divisible by 100.

Now we look at 1! + 2! + 3! + 4! + 5! + 6! + 7! + 8! + 9! mod 100, which is 1 + 2 + 6 + 24 + 20 + 20 +40 +20 + 80 mod 100. 1 + 2 + 6 + 24 + 20 + 20 +40 +20 + 80 = 213 so the remainder is just 13.

if n $\geq$ 10, n! will have the factors 4 and 25, It implies that 100 divide n!. So we only have to calculate the remainder when 1! + 2! + 3! + ... + 9 ! = 409113 is divided by 100. It's easy to see that the answer is 13

Enumerating the factorials from 1 - 9 , we get , 1 ! = 1 2 ! = 2 3 ! = 6 4 ! = 24 5 ! = 120 6 ! = 720 7 ! = 5040 8 ! = 40320 9 ! = 362880

Adding , these factorials only till their tens digits , we get the number 13 , hence remainder is 13 .

For $n\geq 10$ we have that $n!$ is a multiple of $100$. So the remainder is equel to the remainder of $1!+2!+\cdots+9!$ by division by $100$. So the remainder of the first $9$ is $1,2,6,24,20,20,40,20,80$. With sum $213$ so the solution is 13.

100=2²5² then 10!=0 mod 100 , 11!=0 mod 100 ..... 30!=0 mod 100 1! =1 2! =2 3! =6 4! =24 5! =20 6! =20 7! =40 8! =20 9! =80 all mod 100 by adding 13 mod 100

Notice that you only need to focus on the last two digits of each addend since you're working in mod 100. Note that 10!=0 mod 100 (since it has two fives and 8 twos in the factorization) and since 11!,...,30! are all multiples of 10!, they're all equal to 0 mod 100. Thus you only need to consider the sum 1!+2!+....9!=13 mod 100.

for every number,N divided M and left a remainder O

it can be written as $N \equiv O \pmod{M}$

$1! \equiv 1 \pmod{100}$

$2! \equiv 2 \pmod{100}$

$3! \equiv 6 \pmod{100}$

$4! \equiv 24 \pmod{100}$

$5! \equiv 20 \pmod{100}$

$6! \equiv 20 \pmod{100}$

$7! \equiv 40 \pmod{100}$

$8! \equiv 20 \pmod{100}$

$9! \equiv 80 \pmod{100}$

$10! \equiv 0 \pmod{100}$

$11! \equiv 0 \pmod{100}$

$12! \equiv 0 \pmod{100}$

$\ ...$

$30! \equiv 0 \pmod{100}$ $1! + 2! + 3! +\ ... +30! \equiv 213 \pmod{100}$ $1! + 2! + 3! +\ ... +30! \equiv 13 \pmod{100}$

the remainders when dividing by 100 1! = 1 1 x 2 = 2 2 x 3 = 6 6 x 4 = 24 24 x 5 = 20 20 x 6 = 20 20 x 7 = 40 40 x 8 = 20 20 x 9 = 80 80 x 10 = 0 0 x 11 = 0 0 x 30 = 0 sum of remainders = 213 213/100 =2 remainder 13

The first 4 factorials are smaller than 100, so just calculate them the normal way. But 5!=120 is greater than 100, and from here on we will only consider the remainder when divided by 100, so we only care about the 20 in 5!. Now we calculate the remainder of 6!, which can be done by multiplying the remainder of 5! by 6 and omitting every digit except the last two. We continue doing this until 10! where the remainder is 0, it follows that every factorial after 9! is not relevant to the question. So we just sum up the remainders of the first 9 factorials. 1+2+6+24+20+20+40+20+80=213, which leaves us with a remainder of 13.

Reason for neglecting 10! and above is: 10!= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10
so here the set 10,2,5 gives 100 and terms >10! will also have the above expression so terms >= 10! will give remainder 0.

Now for 1!+2!+3!......+9! we can do a bit of calculations !! to simplify calculations eg. 9! = 362880 we can multiply upto 3 places ie. 880 only since remainder of division by hundred is determined upto 3 places

We have to find the remainder when 1! + 2! + 3!......+29!+30! is divided by 100. Since,10! is divisible by 100, n! will be divisible by 100 if n \geq 10. Therefore, n! \geq 10 will not contibute towards the remainder. Hence, only 1! + 2! +3!+4!+5!+6!+7!+8!+9! must be divided by 100. 1! + 2! +3! +4! +5! +6! +7! +8! +9! = 1 +2+ 6+ 24+120 +720 +5040 +40320 +362880 = 409113 When 409113 is divided by 100 the quotient is 4091 and the remainder is 13. Hence, 13 is the remainder.

(1! + 2! + 3! + ... + 29! + 30!) / 100 = 1!/100 + 2!/100 + 3!/100 + ... + 30!/100

10!, 11!, 12! ... 30! are divisible by 100. Hence the sum of these divided by 100 has remainder 0

1!+2!+3!+...+9! = 409113 409113/100 = 4091 R13

So, 1! +2! +3!..+30! =1!+2!...+10! mod 100 So the last digit is 13.(The factorials need not be computed fully,as only the remainder mod 100 is required).

n!≡0 (mod 100) when n>10 because 10!=100 3 4 6 7 8 9 let a=sigma n! from n=1 to n=30 thus a≡sigma n! from n=1 to n=9 use calculator sigma n! from n=1 to n=9=409113≡13 (mod 100) Q.E.D

By calculation, 1!=1, 2!=2, 3!=6, 4!=24. To solve the factorial of higher numbers like 5!, first separate 5! to its factors 5x4x3x2x1. Because of the fact that 4x3x2x1=4!=24, 5!=5(4!)=5(24)=120. These method is used in the next 5 numbers. To know the remainder, we must only take the last 2 digits of every sum. 1!+2!= 3+3!= 9+4!= 33+5!= 53 (taking the last two digits only) +6!= 73+7!= 13+8!= 33+9!= 13+10!= 13. As we continue, 13 appears always in the last two digits of the sum that we get. Therefore, 1!+2!+3!+...+29!+30! is divided by 100, the remainder is 13.

when we compute the last two digits for all the factorials and sum up we will get 13 in the ten's and one's digit at last.thus 13 is the reminder.

I did the summation of x! from 1 to 30. this equals 274410818470142134209703780940313 . the remainder is therefore 274410818470142134209703780940313 mod 100 (is congruent to) 13. Therefore the remainder of (1! + 2! + ... + 30!) when divided by 100 is 13.

we can get remainder when (1factorial+2factorial+.....................+30)/100 by using logic.The factorials of the numbers upto 10 does not end with two zeroes.because they do not have 2,5 twice both as their factors.therefore we can getthe remainder by adding the last two digits of the factorial

10!+11!+.............+13! is divisible by 100 completely with remainder 0. remaining is 1!+2!+............9! = 409113 which when divided by 100 gets remainder as 13