专家免费评估留学案例雅思保分免费备考资料留学热线 400-880-6200

AMC12每日一题(2000年真题#23)

2018-05-21
646 人浏览
分享
收藏

2000 AMC 12竞赛试题/第23题

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 $2^m5^n$ . 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 $2$ s and $5$ s in its factorization, we require larger powers of $5$ than we do 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 $1/4\ \mathrm{(B)}$ .