GreenlightTestPrep wrote:
If N = 1! + 2! + 3! + . . . . . + 12! + 13! + 14! + 15!, what is the remainder when N is divided by 5?
Key concept:
5! is divisible by
5, since 5! = (
5)(4)(3)(2)(1) =
5(some integer)
6! is divisible by
5, since 6! = (6)(
5)(4)(3)(2)(1) =
5(some integer)
7! is divisible by
5, since 7! = (7)(6)(
5)(4)(3)(2)(1) =
5(some integer)
etc...
Given: N = 1! + 2! + 3! + 4! + 5! + 6! + . . . + 12! + 13! + 14! + 15!Substitute to get: N = 1! + 2! + 3! + 4! +
5(some integer) +
5(some integer) + . . .
5(some integer) +
5(some integer)
Factor the last 11 terms: N = 1! + 2! + 3! + 4! +
5(some big integer)
Evaluate the first 4 terms: N = 1 + 2 + 6 + 24 +
5(some big integer)
Simplify: N = 33 +
5(some big integer)
Rewrite 33 as follows: N = 3 + 30 +
5(some big integer)
Factor 5 from the last two terms: N = 3 +
5(6 + (some big integer))
At this point, we can see that N is
3 greater than some multiple of
5.
So, when we divide N by 5, the remainder will be 3
Answer: 3