Question

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 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 Professor Gamble holds the winning ticket?

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$1$

Solution:

The sum of the logarithms is an integer $k$ exactly when the product of the six numbers is $10^k.$ Since $10 = 2 \cdot 5,$ each chosen number must be of the form $2^a 5^b,$ so it comes from $1, 2, 4, 5, 8, 10, 16, 20, 25, 32, 40.$

For each, record the excess of factors of $2$ over factors of $5$ : $0, 1, 2, -1, 3, 0, 4, 1, -2, 5, 2.$ The product is a power of $10$ only if the six chosen values have equal totals of $2$ s and $5$ s, i.e. their excesses sum to $0.$

Working through the possibilities, exactly four valid tickets exist: $\{1, 5, 10, 20, 25, 40\},$ $\{1, 2, 5, 10, 25, 40\},$ $\{1, 2, 4, 5, 10, 25\},$ and $\{1, 4, 5, 10, 20, 25\}.$

Professor Gamble holds one of these four, and only one matches the winning ticket, so the probability is $\dfrac14.$

Thus, the correct answer is B.

2000 AMC 12 Problem 23

Solution:

Problem 23 in Other Years