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Logs puzzle
By Brad Rodgers (P1930) on Tuesday, January 23, 2001 - 08:07 pm:

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 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?

Thanks in advance,

Brad

By Michael Doré (Md285) on Wednesday, January 24, 2001 - 12:50 am:

Try and show that if ai are positive integers:

log a1 + log a2 + ... + log an

is an integer if and only if a1a2...an is a power of ten. (Hint: prove that log a + log b = log ab for any a,b.)

Therefore each of the numbers the professor chooses in the lottery must have only 2 and 5 as prime divisors. Therefore the only numbers it is possible for the professor to have picked are 1,2,4,5,8,10,16,20,25,32,40.

The important thing is that if we multiply 6 of these (our chosen 6) then the number of factors of 2 must be the same as the number of factors of 5 (in the prime factorisation of the product). So all we need to do is consider the difference between the number of 2s and 5s in the factorisation in each of these numbers. So if we look at this difference:

(0),(1),(2),(-1),(3),(0),(4),(1),(-2),(5),(2)

So we just want to pick 6 of these that add to 0.

Now try and prove the following:

i) The (4) can be thrown out - it can never be used. Likewise the (5) and the (3).

ii) The (-1) and (-2) must be both used.

From there you just need to calculate the number of possibilities.

By Tom Hardcastle (P2477) on Wednesday, January 24, 2001 - 12:52 am:

An important property of logarithms is that log A + log B = log AB.
So log A + log B + log C + log D + log E + log F = log ABCDEF = k where k is some integer.
So ABCDEF = 10k.
You should then be able to find all possible values for A, B etc. by considering the factors of 10k.

Tom.

By Brad Rodgers (P1930) on Wednesday, January 24, 2001 - 06:12 pm:

Using your hints, I've been able to get the answer 1/4.

Thanks, you've been a great help,

Brad