It's easier not to multiply everything out.
How many of these numbers:
1
!
,
2
!
,
3
!
,
4
!
,
…
,
2
0
1
5
!
are even numbers?
Image Credit: Flickr no_1741 .
The answer is 2014.
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6 solutions
Every factorial number but 1 and 0 is multiplied by two, and there we have from 1! to 2015! so 2014 of them are even numbers.
1 ! = 1 2 ! = 1 × 2 = 2 3 ! = 1 × 2 × 3 = 6 4 ! = 1 × 2 × 3 × 4 = 2 4 . n ! = 1 × 2 × 3 × 4 × . . . . n ! . . 2 0 1 5 ! = 1 × 2 × 3 × 4 × . . . . . × 2 0 1 5 !
For every n ! where n > 1 , they are a multiple of 2, which makes them even.
So, the total number of even numbers between 1 ! & 2 0 1 5 ! will be, 2 0 1 5 − 1 = 2 0 1 4 .
Moderator note:
This is not a valid solution. You have only shown that 2 ! , 3 ! and 4 ! are even numbers but you didn't show that 5 ! , 6 ! , … , 2 0 1 5 ! are even numbers as well.
The keyword here is to mention that 2 divides 2 ! , 3 ! , … , 2 0 1 5 ! , so these numbers are all even number.
What's wrong in the solution?
Isn't "2 divides these numbers" & "these numbers are multiple of 2" are same statements?
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@Calvin Lin Didn't notice the note before. Can you please explain what's wrong there?
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As stated, you have only shown it for 2 ! , 3 ! , 4 ! . Why is it true for 5!? What is the logic behind it?
Once again, not all of us are mind readers, and we do not know exactly what you are thinking. We can only read what you have written, to decide if what you say is true.
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@Calvin Lin – I have shown first 4 numbers as an example. Then said, "Every factorial (given in the question) is a multiple of 2 except 1!".
Doesn't it make the answer clear enough? And how can it not be valid??
However, I'm editing the solution slightly for easier understanding.
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@Md Omur Faruque – Right. It now follows that you are doing a solution by induction, or a solution of "listing out all possible cases".
previously, what you had was merely "listing out 4 cases".
1!=1 it's an odd.. Next 2 is even number... even number multiplied by even or odd number gives results even number only... After 1! Each term has 2... so All other terms are even numbers... Hence answer was 2015-1=2014
All of them are going to end up even because you multiply by 2 in the end. Ex: 5! is 5x4x3x2*1. Any number times 2 is even. The exception is 1! Which is just 1.
5! and so on end in zero (120, 720, 5040 etc,) as well as the number two in their factorization. This is also true for 2!, 3! and 4!. The only number that does not have two in their factorization. Henceforth, their are 2014 even numbers in 1!-2015!
Apart from 1 ! = 1 which is odd , 2 ! , 3 ! , 4 ! , … 2 0 1 5 ! have at least one power of 2 in their factorization , making them even. Hence , the even numbers are 2 0 1 5 − 1 = 2 0 1 4 .