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
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
To find the probability that Joe holds the winning ticket, we need to determine the total number of possible tickets and the number of tickets that satisfy the given condition.
The total number of possible tickets is the number of ways to choose 6 different integers from a set of 46. This can be calculated using the combination formula:
C(46, 6) = 46! / (6! * (46-6)!) = 9,366,819
To determine the number of tickets that satisfy the given condition, we need to consider the possible sums of the base-ten logarithms that yield an integer.
The sum of the base-ten logarithms of a set of numbers is equal to the base-ten logarithm of the product of those numbers. So, we need to find sets of 6 integers whose product is a perfect power of 10.
Let's analyze the product of numbers from 1 through 9:
P(1, 2, 3, 4, 5, 6) = 720
P(1, 2, 3, 4, 5, 7) = 5040 = 10^3.70
P(1, 2, 3, 4, 5, 8) = 40,320 = 10^4.60
P(1, 2, 3, 4, 5, 9) = 362,880 = 10^5.56
We can see that when the product of the numbers has an exponent that is a multiple of 10, the sum of the base-ten logarithms will be an integer.
Therefore, there are 4 sets of 6 numbers that satisfy the condition: (1, 2, 3, 4, 5, 6), (1, 2, 3, 4, 5, 7), (1, 2, 3, 4, 5, 8), and (1, 2, 3, 4, 5, 9).
The probability that Joe holds the winning ticket is the number of winning tickets divided by the total number of possible tickets:
P(winning) = 4 / 9,366,819
Simplifying this fraction, we find:
P(winning) = 1 / 2,341,705.25
Therefore, the answer is approximately 1 in 2,341,705.25.
The nearest answer option is (E) 1, as the probability is extremely low.