The log of some number $x$ sum would be equal to its height from the point $10^0$. Thus, we want the same amount of space from the ranges $[1-10),[10-100),[100-1000),$ $[1000-10000),[10000-100000)$ as we do from $[100000,1000000)$ to $[1000000-10000000)$. Since these all have the same range in powers of $10$, we need to find all combinations of six numbers where three are in $[100000,1000000)$ and three are in $[1000000,10000000)$.
$\dbinom{6}{3} = 20$, and the total number of combinations for 6 numbers from a set of 46 is $\dbinom{46}{6} = 9,366,819$
Thus, the probability is $\boxed{\frac{1}{5}}$.
- AlexLikeMath
Joe 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 of his six numbers is an 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 that Joe holds the winning ticket?
A) 1/5
B) 1/4
C) 1/3
D) 1/2
E) 1
1 answer