I will prove that the largest power of 2 that divides (n+1)*(n+2)*...*2n is 2^n for all n.
For n=1, the product is 2. The largest power of 2 that divides 2 is 2^1. Therefore, it is true for n=1.
Suppose it is true for n. Then, the largest power of 2 that divides (n+1)*(n+2)*...*2n is 2^n. Let f(n)=(n+1)*(n+2)*...*2n. To get f(n+1)=(n+2)*(n+3)*...*(2n+2), we divide f(n) by n+1 and multiply it by 2n+1 and 2n+2. Dividing it by n+1 and multiplying it by 2n+2 is the same as multiplying it by 2. That gives one more power of 2. The number 2n+1 is odd, so it gives no powers of 2. Therefore, we have one more power of 2. Since the largest power of 2 that divides f(n) is 2^n, the largest power of 2 that divides f(n+1) is 2^(n+1). Therefore, if it true for n, then it is true for n+1. By induction, it is true for all n.
Therefore, the largest power of 2 that divides 2012*2013*...*4022 is 2^2011. Since 2011 is odd, the largest power of 2 that divides 2011*2012*2013*...*4022 is also 2^2011. Therefore, the answer is 2011.

Posted by Math Man
on 20111206 19:41:31 