Problem:
Define n!! to be n(n−2)(n−4)…3⋅1 for n odd and n(n−2)(n−4)…4⋅2 for n even. When ∑i=12009​(2i)!!(2i−1)!!​ is expressed as a fraction in lowest terms, its denominator is 2ab with b odd. Find 10ab​.
Solution:
The fact that (m2m​)=m!m!2mm!(2m−1)!!​=m!2m(2m−1)!!​ is an integer implies that if pa is a power of an odd prime dividing (2m)!!, then pa divides m! and hence (2m−1)!!. Thus any odd prime powers which divide the denominator in any given term in the sum divide the numerator as well. Therefore, when reduced to lowest terms, the denominator of the m th term in the sum is the highest power of 2 that divides (2m)!!. Hence 2ab is the highest power of 2 dividing (2⋅2009)!!=4018!!. Also note that b=1 because b is odd and 2ab is a power of 2. Because 4018!!=22009⋅2009!, the highest power of 2 that divides 4018!! is
2009+⌊22009​⌋+⌊222009​⌋+⋯+⌊2102009​⌋2009+1004+502+251+125+62+31+15+7+3+1​==4010.​
Thus the requested answer is 104010⋅1​=401​.
The problems on this page are the property of the MAA's American Mathematics Competitions