# plz help to solve

• April 18th 2008, 07:49 PM
mythri sharma
plz help to solve
find the remainder when 1!+2!+3!+..........99!+100! divided by 12.
• April 18th 2008, 07:55 PM
Mathstud28
Quote:

Originally Posted by mythri sharma
find the remainder when 1!+2!+3!+..........99!+100! divided by 12.

I am not positive...but would you apply Wilson's Theorem?
• April 18th 2008, 08:34 PM
awkward
Quote:

Originally Posted by mythri sharma
find the remainder when 1!+2!+3!+..........99!+100! divided by 12.

Hint: n! is evenly divisible by 12 for all n > 3.
• April 19th 2008, 01:28 AM
Isomorphism
Quote:

Originally Posted by mythri sharma
find the remainder when 1!+2!+3!+..........99!+100! divided by 12.

Try writing the factorials as a product.

Now 12 = 4 x 3

1! = 1
2! = 2 x 1
3! = 3 x 2 x 1
4! = (4 x 3) x 2 x 1 <------ There is a 4 x 3 here
5! = 5 x (4 x 3) x 2 x 1 <------ There is a 4 x 3 here
6! = 6 x 5 x (4 x 3) x 2 x 1 <------ There is a 4 x 3 here
7! = 7 x 6 x 5 x (4 x 3) x 2 x 1 <------ There is a 4 x 3 here

Well I am sure you can see by now that 12 will divide any factorial greater than or equal to 4. That means the only fellows who can leave remainders are 1!+2!+3! = 1+2+6 = 9. So 9 is the remainder when divided by 12.

Now can you try a related question:
"Which is the largest factorial number that is not divisible by 27?"

Try it :D
• April 19th 2008, 04:35 PM
Mathstud28
Quote:

Originally Posted by Isomorphism
Try writing the factorials as a product.

Now 12 = 4 x 3

1! = 1
2! = 2 x 1
3! = 3 x 2 x 1
4! = (4 x 3) x 2 x 1 <------ There is a 4 x 3 here
5! = 5 x (4 x 3) x 2 x 1 <------ There is a 4 x 3 here
6! = 6 x 5 x (4 x 3) x 2 x 1 <------ There is a 4 x 3 here
7! = 7 x 6 x 5 x (4 x 3) x 2 x 1 <------ There is a 4 x 3 here

Well I am sure you can see by now that 12 will divide any factorial greater than or equal to 4. That means the only fellows who can leave remainders are 1!+2!+3! = 1+2+6 = 9. So 9 is the remainder when divided by 12.

Now can you try a related question:
"Which is the largest factorial number that is not divisible by 27?"

Try it :D

Would it be $8!$ because since the only numbers(that you can't have a repeated factor) that multiply to 27 are 9*3 which only occurs first in $9!$?
• April 20th 2008, 06:10 AM
Isomorphism
Quote:

Originally Posted by Mathstud28
Would it be $8!$ because since the only numbers(that you can't have a repeated factor) that multiply to 27 are 9*3 which only occurs first in $9!$?

Yes, right :D
• April 20th 2008, 06:35 AM
Mathstud28
Quote:

Originally Posted by Isomorphism
Yes, right :D

Hooray!