How to solve factorial related problems

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In summary, the conversation discusses a problem involving factorials and the general format for solving it. The problem is to find the tens digit in the sum of several factorials, and it is solved by realizing that any factorial greater than or equal to 10 will not contribute to the tens digit. The final answer is 4. The conversation also mentions the use of number theory to solve similar problems.
  • #1
Thundagere
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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!
 
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  • #2
Thundagere said:
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?
 
  • #3
Isn't every factorial above a certain number a multiple of 100, and therefore wouldn't add to the tens digit?
 
  • #4
Char. Limit said:
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 :)
 
  • #5
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!
 

1. What is factorial and how does it work?

Factorial is a mathematical operation denoted by the symbol "!". It is used to calculate the product of all positive integers from 1 to a given number. For example, 5 factorial (5!) is equal to 1 x 2 x 3 x 4 x 5 = 120. In other words, it is the number of ways to arrange a set of objects in a specific order.

2. How do I solve a factorial problem?

To solve a factorial problem, you first need to understand the given problem and identify the number for which you need to find the factorial. Then, you can use a calculator or manually calculate the factorial by multiplying all the positive integers from 1 to the given number.

3. Can I use factorial for non-integer numbers?

No, factorial can only be used for positive integers. It is not defined for non-integer numbers or negative numbers.

4. How can I use factorial in probability problems?

Factorial is commonly used in probability problems to calculate the number of possible outcomes. For example, if you have 5 objects and you want to know how many different ways they can be arranged, you can calculate 5 factorial (5!) to get the total number of possible outcomes.

5. Are there any shortcuts for solving factorial problems?

Yes, there are some shortcuts that can be used to solve factorial problems. For example, if the given number is a factorial itself, you can use the inverse factorial function (x!)-1 to find the original number. You can also use the factorial formula nPr = n! / (n-r)! to find the number of permutations for a given set of objects.

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