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Re: If N = 1! + 2! + 3! + . . . . . + 12! + 13! + 14! + 15!, what is the r [#permalink]
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So it will boils down to 4! + 3! + 2! + 1 i.e. 4*3*2 + 3*2 + 2 + 1 ---> 24 + 6 + 2 + 1 ----> 33

Hence reminder will be 3
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Re: If N = 1! + 2! + 3! + . . . . . + 12! + 13! + 14! + 15!, what is the r [#permalink]
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GreenlightTestPrep wrote:
If N = 1! + 2! + 3! + . . . . . + 12! + 13! + 14! + 15!, what is the remainder when N is divided by 5?


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3


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
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Re: If N = 1! + 2! + 3! + . . . . . + 12! + 13! + 14! + 15!, what is the r [#permalink]
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Given that N = 1! + 2! + 3! + . . . . . + 12! + 13! + 14! + 15! and we need to find the remainder when N is divided by 5

In these type of problems we need to find a pattern and then try to eliminate the terms based on the pattern.

Now, N = 1! + 2! + 3! + . . . + 15! and we know that a higher factorial will always be a multiple of lower factorial
Ex: 4! = 1*2*3*4 = 3! * 4

So, if we can prove that the remainder of lets say n! with 5 is 0 then remainder of all higher factorials like (n+1)!, (n+2)! will be 0 too.
So, lets start finding the remainder of factorials from 1 onwards by 5 and see when we can reach a remainder of 0

Remainder of 1! by 5 = Remainder of 1 by 5 = 1
Remainder of 2! by 5 = Remainder of 1*2(=2) by 5 = 2
Remainder of 3! by 5 = Remainder of 1*2*3(=6) by 5 = 1
Remainder of 4! by 5 = Remainder of 1*2*3*4(=24) by 5 = 4
Remainder of 5! by 5 = Remainder of 1*2*3*4*5 by 5 = 0 (As it is a multiple of 5)
Now, all factorials from 5! onwards will give a remainder of 0 when divided by 5, as all of them will be a multiple of 5.

=> Remainder of N with 5 = Remainder of (1! + 2! + 3! + . . . + 15!) by 5 = Remainder of 1! by 5 + Remainder of 2! by 5 and so on
(Theory: Remainder of sum of two numbers by a number is same as the sum of remainder of those two numbers (individually) by that number)

= 1 + 2 + 1 + 4 + 0....+0 = 1 + 2 + 1 + 4 = 8
But remainder of N by 5 cannot be greater than 4
(Theory: If we are dividing a number by n then we can get remainders only from 0 to n-1)

=> Remainder will be = Remainder of 8 by 5 = 3

So, Answer will be 3
Hope it helps!

Watch the following video to learn the Basics of Remainders

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Re: If N = 1! + 2! + 3! + . . . . . + 12! + 13! + 14! + 15!, what is the r [#permalink]
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All of the factorials 5! and up are multiples of 5, so that whole block is a multiple of 5 plus + 24 + 6 + 2 + 1, which gives you 3 more than a multiple of 5
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