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考题15-16 2015 AMC 10B

2018-08-06 重点归纳

AMC10数学竞赛是美国高中数学竞赛中的一项，是针对高中一年级及初中三年级学生的数学测试，该竞赛开始于2000年，分A赛和B赛，于每年的2月初和2月中举行，学生可任选参加一项即可。不管是对高校申请还是今后在数学领域的发展都极其有利！那么接下来跟随小编来看一下AMC10的官方真题以及官方解答吧：

Problem

The town of Hamlet has $3$ people for each horse, $4$ sheep for each cow, and $3$ ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?

$考题15-16 2015 AMC 10B$

Solution 15

Solution 1

Let the amount of people be $p$ , horses be $h$ , sheep be $s$ , cows be $c$ , and ducks be $d$ . We know

$h=3p$ $c=4s$ $p=3d$

Then the total amount of people, horses, sheep, cows, and ducks may be written as $3p+p+4s+s+9d=13d+5s$ . Looking through the options, we see $47$ is impossible to make for integer values of $h$ and $c$ . So the answer is $\boxed{\textbf{(B)} 47}$ .

Solution 2

As the solution above says, the total amount of people, horses, sheep, cows, and ducks may be written as $13d+5s$ . However, instead of going through each of the solutions and testing the the options, you can use the Chicken McNugget Theorem to find the greatest number of people, horses, sheep, cows, and ducks that cannot be written in the form $13d+5s$ .

so our answer is $\boxed{\textbf{(B)} 47}$ .

Problem 16

Al, Bill, and Cal will each randomly be assigned a whole number from $1$ to $10$ , inclusive, with no two of them getting the same number. What is the probability that Al's number will be a whole number multiple of Bill's and Bill's number will be a whole number multiple of Cal's? $考题15-16 2015 AMC 10B$

Solution

We can solve this problem with a brute force approach.

If Cal's number is $1$ :

If Bill's number is $2$ , Al's can be any of $4, 6, 8, 10$ .
If Bill's number is $3$ , Al's can be any of $6, 9$ .
If Bill's number is $4$ , Al's can be $8$ .
If Bill's number is $5$ , Al's can be $10$ .
Otherwise, Al's number could not be a whole number multiple of Bill's.

If Cal's number is $2$ :

If Bill's number is $4$ , Al's can be $8$ .
Otherwise, Al's number could not be a whole number multiple of Bill's while Bill's number is still a whole number multiple of Cal's.

Otherwise, Bill's number must be greater than $5$ , i.e. Al's number could not be a whole number multiple of Bill's.

Clearly, there are exactly $9$ cases where Al's number will be a whole number multiple of Bill's and Bill's number will be a whole number multiple of Cal's. Since there are $10*9*8$ possible permutations of the numbers Al, Bill, and Cal were assigned, the probability that this is true is $\frac9{10*9*8}=\boxed{\text{(\textbf C) }\frac1{80}}$