How to solve factorial related problems

[Thundagere]

Hi all,
I know what factorials are, obviously, and permutations and combinations, but what I Don't know is, given a problem with factorials in it, is there a general format for solving?
For instance, an old AMC 10 problem:

10B-#11. What is the tens digit in the sum 7! + 8! + 9! +
. . . + 2006! ?
(A) 1 (B) 3 (C) 4 (D) 6 (E) 9

How would one go about solving this?
All help appreciated!

 

[chiro]

Hi all,
I know what factorials are, obviously, and permutations and combinations, but what I Don't know is, given a problem with factorials in it, is there a general format for solving?
For instance, an old AMC 10 problem:

10B-#11. What is the tens digit in the sum 7! + 8! + 9! +
. . . + 2006! ?
(A) 1 (B) 3 (C) 4 (D) 6 (E) 9

How would one go about solving this?
All help appreciated!

Hey Thundagere and welcome to the forums.

This reminds me of a number theory problem.

For finding the final digit of the number (i.e. the lowest ranked digit in the number), then this reduces to finding the number N (mod 10).

Now based on this as well as other congruence arithmetic identities, do you have any new ideas that you can use to solve this?

 

[Char. Limit]

Isn't every factorial above a certain number a multiple of 100, and therefore wouldn't add to the tens digit?

 

[chiro]

Isn't every factorial above a certain number a multiple of 100, and therefore wouldn't add to the tens digit?

That's a great observation and makes the problem really easy :)

 

[Thundagere]

OK
So anything greater than or equal to 10 we can discount, since 2 * 5 * 10 is 100, which wouldn't matter.
So with that in mind, we're looking at
7! + 8! + 9!
=7!(1 + 8 + 8 * 9)
=5040(81)
=408,240
So it's then 4?
For these types of AMC 10 problems, could reasoning it out as above suffice? I'm not well versed in number theory, so...yeah.
Thanks for your help!