- #1
davon806
- 148
- 1
Homework Statement
Hi,I found this problem when I was reading a book,I knew the answer but there is one thing that I don't understand.Here is the question:
Given C(2015,m),find the smaller of m such that C(2015,m) is an even number
Homework Equations
The Attempt at a Solution
Here is the answer:(sorry for being quite messy)
Expand C(2015,m),which gives:
(2015)(2014)... (numerator)
m(m-1)(m-2)...(4)(3)(2)(1)...(2015-m)(2015-m+1)... (denominator)
After some arrangement it becomes:
(2015)(2014)...(2015-(m-1))
m(m-1)...(4)(3)(2)(1)
which is :
(2015)(2014)...(2016-m)
(1)(2)(3)(4)...(m)
In my book,the answer said the first even number occurs when (2016-m)/m is an even number
Therefore 2016/m - 1 must be even,then 2016/m must be odd.
Since 2016 = 2^5 * 63,the smallest number of m is 2^5 = 32
My question:
I know the answer is correct.However,let's just expand a few terms up to 2011 (i.e.m=5)
(2015)(2014)(2013)(2012)(2011)
(5) (4) (3) (2) (1)
which gives:
(403)(1007)(671)(503)(2011) = 2.75*(10^14)
Since m is small,the final product C(2015,m) has a high chance to be an integer as the factors in the denominator is quite small.
However,how can we ensure this is still the case when m is large(Just say m=32 given in my case),could it be a fraction instead of an even number?There might be terms like (1984/19) which makes the final term neither odd nor even.
Thanks very much.