Japan Junior Mathematical Olympiad

For a real number $x$, denote by $\lfloor x\rfloor$ the greatest integer $\le x$. Then, for every positive integer $n$, there are exactly $\lfloor \frac{n}{2}\rfloor$ even numbers among $1,2,\dots ,n$ and therefore, $1\times 2\times 3\times \cdots \times n$ can be divided by $2$ at least $\lfloor \frac{n}{2}\rfloor$ times, and if we go through the division by $2$ $\lfloor \frac{n}{2}\rfloor$ times, then you get as the quotient $1\times 2\times \cdots \times \lfloor \frac{n}{2}\rfloor \times q$, where $q$ is a product of odd integers. Similarly, $1\times 2\times \cdots \times \lfloor \frac{n}{2}\rfloor$ can be divided by $2$ at least $\lfloor \frac{\lfloor \frac{n}{2}\rfloor }{2}\rfloor =\lfloor \frac{n}{4}\rfloor$ times, and if we carry through the division by $2$ so many times you get the quotient $1\times 2\times \dots \lfloor \frac{n}{2}\rfloor \times q^{\prime }$, where $q^{\prime }$ is a product of odd integers. If you repeat this procedure $k-1$ times, where $k$ is the smallest positive integer for which $2^k>n$, then we see that $1\times 2\times 3\times \cdots \times n$ can be divided by $2$ exactly $\lfloor \frac{n}{2}\rfloor +\lfloor \frac{n}{4}\rfloor +\lfloor \frac{n}{8}\rfloor +\cdots +\lfloor \frac{n}{2^k}\rfloor$ times, since after you carry through the division by $2$ so many times you end up as the quotient with a number which is a product of odd integers only.

Since $2^{10}=1024<2008<2028=2^{11}$, and $2^9=512<1003<1024=2^{10}$, we can conclude from the discussion above by letting $n=2008$ and $1003$, respectively, that $1\times 2\times \cdots \times 2008$ can be divided by $2$ exactly $\lfloor \frac{2008}{2}\rfloor +\lfloor \frac{2008}{4}\rfloor +\lfloor \frac{2008}{8}\rfloor +\cdots +\lfloor \frac{2008}{1024}\rfloor$ times and that $1\times 2\times \cdots \times 1003$ can be divided by $2$ exactly $\lfloor \frac{1003}{2}\rfloor +\lfloor \frac{1003}{4}\rfloor +\lfloor \frac{1003}{8}\rfloor +\cdots +\lfloor \frac{1003}{512}\rfloor$ times. Since the desired answer is the difference of these two numbers, we get $$(1004+502+251+125+62+31+15+7+3+1)$$

$-(501+250+125+62+31+15+7+3+1)=1006.$ $$Alternatively,since$$ $$\begin{array}{rl}1004\times 1005\times \cdots \times 2008& =\frac{1\times 2\times 3\times 4\times \cdots \times 2008}{1\times 2\times \cdots \times 1003}\\ & =1\times 3\times 5\times \cdots \times 2007\times \frac{2\times 4\times 6\times \cdots \times 2006}{1\times 2\times 3\times \cdots \times 1003}\times 2008\\ & =1\times 3\times 5\times \cdots \times 2007\times 2^{1003}\times (2^3\times 251)\\ & =2^{1006}\times 1\times 3\times 5\times \cdots \times 2007\times 251,\end{array}$$ $$fromwhichweconcludethatthegivennumbercanbedividedby$2$exactly$1006$ times.