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2000 AMC 12 Problems/Problem 23

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Problem

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?

\text {(A)}\ 1/5 \qquad \text {(B)}\ 1/4 \qquad \text {(C)}\ 1/3 \qquad \text {(D)}\ 1/2 \qquad \text {(E)}\ 1

Solution

The product of the numbers have to be a power of 10 in order to have an integer base ten logarithm. Thus all of the numbers must be in the form 2^m5^m. Listing out such numbers from 1 to 46, we find 1,2,4,5,8,10,16,20,25,32,40 are the only such numbers. Immediately it should be noticed that there are a larger number of powers of 2 than of 5. Since a number in the form of 10^k must have the same number of 2s and 5s in its factorization, we require larger powers of 5 of those of 2. To see this, for each number subtract the power of 5 from the power of 2. This yields 0,1,2,-1,3,0,4,1,-2,5,2, and indeed the only non-positive terms are 0,0,-1,-2. Since there are only two zeros, the largest number that Professor Gamble could have picked would be 2.

Thus Gamble picks numbers which fit -2 + -1 + 0 + 0 + 1 + 2, with the first four having already been determined to be \{25,5,1,10\}. The choices for the 1 include \{2,20\} and the choices for the 2 include \{4,40\}. Together these give four possible tickets, which makes Professor Gamble’s probability 1/4\ \mathrm{(B)}.

See also

2000 AMC 12 (ProblemsResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
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