Professor Gamble buys a lottery ticket, which requires that he pick six different integers from $1$ through $46$, inclusive. He chooses his numbers so that the sum of the base-ten [logarithms](https://artofproblemsolving.com/wiki/index.php/Logarithm) of his six numbers is an [integer](https://artofproblemsolving.com/wiki/index.php/Integer). It so happens that the integers on the winning ticket have the same property— the sum of the base-ten logarithms is an integer. What is the [probability](https://artofproblemsolving.com/wiki/index.php/Probability) that Professor Gamble holds the winning ticket?
$\textbf {(A)}\ 1/5 \qquad \textbf {(B)}\ 1/4 \qquad \textbf {(C)}\ 1/3 \qquad \textbf {(D)}\ 1/2 \qquad \textbf {(E)}\ 1$
The product of the numbers has to be a power of $10$ in order to have an integer base ten logarithm. Thus all of the numbers must be in the form $2^m5^n$. Listing out such numbers from $1$ to $46$, we find $1,2,4,5,8,10,16,20,25,32,40$ are the only such numbers. Immediately it should be noticed that there are a larger number of powers of $2$ than of $5$. Since a number in the form of $10^k$ must have the same number of $2$s and $5$s in its factorization, we require larger powers of $5$ than we do of $2$. To see this, for each number subtract the power of $5$ from the power of $2$. This yields $0,1,2,-1,3,0,4,1,-2,5,2$, and indeed the only non-positive terms are $0,0,-1,-2$. Since there are only two zeros, the largest number that Professor Gamble could have picked would be $2$. 
Thus Gamble picks numbers which fit $-2 + -1 + 0 + 0 + 1 + 2$, with the first four having already been determined to be $\{25,5,1,10\}$. The choices for the $1$ include $\{2,20\}$ and the choices for the $2$ include $\{4,40\}$. Together these give four possible tickets, which makes Professor Gamble’s probability $1/4\ \mathrm{(B)}$.