Gavin buys a lottery ticket, and in accordance with the rules he picks six different integers from through 1 to 46 inclusively. He chooses his numbers so that the sum of the base-ten logarithms of his six numbers is an integer.

By coincidence, the integers on the winning ticket have the same property - that is: the sum of the base-ten logarithms of the six numbers is an integer.

What is the probability that Gavin holds the winning ticket?

Consider:

(1,10), (2,5), (4,25), (5,20)

Erasing one of the above pairs, you are left with 6 numbers, that when multiplied yield 10^n (integer n).

Since there are 4 possibilities of selecting 6 numbers that qualify and Gavin selected one of them.

**The probability that Gavin holds the winning ticket is 1/4=.25.**

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oops: 5 appears twice..**I g n o r e the above:**

__The corrected solution:__

**1 2 4 5 10 25**

**1 2 5 10 25 40**

**1 4 5 10 20 25 **

**1 5 10 20 25 ****40**

**The probability that Gavin holds the winning ticket**

** is 1/4= 25%**

__Same result, but now due to correct reasoning.__

*Edited on ***October 6, 2014, 4:11 pm**