# Probability contest problem

• Jan 10th 2010, 10:05 PM
Vicky1997
Probability contest problem
Professsor 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 winner ticket?

The answer is 1/4 and I got it after an hour of guessing and checking.
I summarized how I reached the answer, but I wanted to check if it is reasonable. I'm starting to see my solutions are too sloppy, and they could be wrong even though I got the correct answer. Could someone show me a way to solve this with less guessing?

Vicky.
• Jan 11th 2010, 10:28 AM
Soroban
Hello, Vicky!

Quote:

Professsor Gamble buys a lottery ticket, which requires that he pick six different integers from 1 -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 winner ticket?

Suppose the winning numbers are: . $a,b,c,d,e,f$

The sum of the six logs is an integer: . $\log a + \log b + \log c + \log d + \log e + \log f \:=\:k$

This means that the log of the product of the numbers is an integer: . $\log(abcde\!f) \:=\:k$

. . Hence: . $abcde\!f \:=\:10^k$

That is, the product of the 6 different numbers is a power of 10.

I found only four cases:

. . $\begin{array}{ccc}1\cdot2\cdot4\cdot5\cdot10\cdot2 5 &=& 10^4 \\
1\cdot2\cdot5\cdot10\cdot25\cdot40 &=&10^5 \\
1\cdot4\cdot5\cdot10\cdot20\cdot25 &=& 10^5 \\
1\cdot5\cdot10\cdot20\cdot25\cdot40 &=& 10^6
\end{array}$

Therefore, the Professor's probability of winning is $\tfrac{1}{4}$