# Winning ticket (Challenging)

• April 18th 2010, 05:40 AM
acc100jt
Winning ticket (Challenging)
Professor Gamble buys a lottery ticket, which requires that he pick six different integers from 1 to 46 inclusive. He chooses his numbers so that the sum of the base-ten logarithm 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?
• April 18th 2010, 08:24 AM
Soroban
Hello, acc100jt!

A fascinating problem!
I have a game plan and a weak finish.

Well, maybe not . . .

Quote:

Professor Gamble bought a lottery ticket; he picked six different integers from 1 to 46 inclusive.
He chose his numbers so that the sum of the base-ten logs 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?

Let his six numbers be: . $a,b,c,d,e,f.$

If the sum of the logs of his six numbers is an integer,
. . then the product of the 6 numbers is a power-of-10.

$\log a + \log b + \log c + \log d + \log e + \log f \:=\:n \quad\Rightarrow\quad \log(a\!\cdot\!b\!\cdot\!c\!\cdot\!d\!\cdot\!e\!\c dot\!f) \:=\:n$

. . Hence: . $a\!\cdot\!b\!\cdot\!c\!\cdot\!d\!\cdot\!e\!\cdot\! f \:=\:10^n$

His product ranges from a minimum of: . $1\!\cdot\!2\!\cdot\!3\!\cdot\!4\!\cdot\!5\!\cdot\! 6 \:=\:720$

. . to a maximum of: . $41\!\cdot\!42\!\cdot\!43\!\cdot\!44\!\cdot\!45\!\c dot\!46 \:=\:6,\!744,\!109,\!680$

Hence, his product is one of: . $10^3,\;10^4,\;10^5\;\hdots \;10^9$

. . We have only 7 cases to consider.

Can his product be $10^3 = 1000$? . . . No

$1000 \;=\;1\cdot2\cdot2\cdot2\cdot5\cdot5\cdot5$

. . which cannot be partitioned into 6 distinct factors.

Can his product be $10^4 = 10,\!000$? . . . Yes

$10,\!000 \;=\;1\cdot2\cdot2\cdot2\cdot2\cdot5\cdot5\cdot5\c dot5$

. . which factors into: . $1\cdot2\cdot4\cdot5\cdot10\cdot25$

Can his product be $10^5 = 100,\!000$? . . . Yes

$100,\!000 \;=\;1\cdot2\cdot2\cdot2\cdot2\cdot2\cdot5\cdot5\c dot5\cdot5\cdot5$

. . which factors into: . $1\cdot4\cdot5\cdot10\cdot20\cdot25$

Can his product be $10^6 = 1,\!000,\!000$? . . . Yes

$1,\!000,\!000 \;=\;1\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot5\c dot5\cdot5\cdot5\cdot5\cdot5$

. . which factors into: . $1\cdot5\cdot10\cdot20\cdot25\cdot40$

Here's where it gets interesting.

Can his product be $10^7 = 10,\!000,\!000$ or larger? . . . No

$10,\!000,\!000 \;=\;1\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\: \cdot\:\underbrace{5\cdot5\cdot5\cdot5\cdot5\cdot5 \cdot5}_{\text{seven factors of 5}}$

The only numbers from 1 to 46 of the form . $2^m5^n$

. . are: . $1, 2, 4, 5, 8, 10, 16, 20, 25, 32, 40$

The ones containing factors-of-5 are: . $\begin{Bmatrix} 5 &=& 5 \\ 10 &=& 2\cdot 5 \\ 20 &=& 2^2\cdot5 \\ 25 &=& 5^2 \\ 40 &=& 2^3\cdot5 \end{Bmatrix}$

Using all of them, we can account for six factors-of-5 . . . and no more.

Hence. $10^7$ cannot be factored into six distinct factors from 1 to 46.

. . The same holds true for: . $10^8\,\text{ and }\,10^9$

I found three outcomes that satisfy the log/sum/integer constraint.

Therefore, the probability that Professor Gamble wins is $\frac{1}{3}$

• April 18th 2010, 06:16 PM
acc100jt
Brilliant!!! Thanks a lot!