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AMC数学竞赛真题2017年10A 19-20

2018-08-27 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数学竞赛真题以及官方解答吧：

Problem 19

Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of 5 chairs under these conditions?

$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 40$

Solution 1

For notation purposes, let Alice be A, Bob be B, Carla be C, Derek be D, and Eric be E. We can split this problem up into two cases:

$\textbf{Case 1: }$ A sits on an edge seat.

Then, since B and C can't sit next to A, that must mean either D or E sits next to A. After we pick either D or E, then either B or C must sit next to D/E. Then, we can arrange the two remaining people in two ways. Since there are two different edge seats that A can sit in, there are a total of $2 \cdot 2 \cdot 2 \cdot 2 = 16$ .

$\textbf{Case 2: }$ A does not sit in an edge seat.

In this case, then only two people that can sit next to A are D and E, and there are two ways to permute them, and this also handles the restriction that D can't sit next to E. Then, there are two ways to arrange B and C, the remaining people. However, there are three initial seats that A can sit in, so there are $3 \cdot 2 \cdot 2 = 12$ seatings in this case.

Adding up all the cases, we have $AMC10数学竞赛真题$ .

Solution 2

Label the seats $1$ through $5$ . The number of ways to seat Derek and Eric in the five seats with no restrictions is $5*4=20$ . The number of ways to seat Derek and Eric such that they sit next to each other is $8$ (which can be figure out quickly), so the number of ways such that Derek and Eric don't sit next to each other is $20-8=12$ . Note that once Derek and Eric are seated, there are three cases.

The first case is that they sit at each end. There are two ways to seat Derek and Eric. But this is impossible because then Alice, Bob, and Carla would have to sit in some order in the middle three seats which would lead to Alice sitting next to Bob or Carla, a contradiction. So this case gives us $0$ ways.

Another possible case is if Derek and Eric seat in seats $2$ and $4$ in some order. There are 2 possible ways to seat Derek and Eric like this. This leaves Alice, Bob, and Carla to sit in any order in the remaining three seats. Since no two of these three seats are consecutive, there are $3!=6$ ways to do this. So the second case gives us $2*6=12$ total ways for the second case.

The last case is if once Derek and Eric are seated, exactly one pair of consecutive seats is available. There are $12-2-2=8$ ways to seat Derek and Eric like this. Once they are seated like this, Alice cannot not sit in one of the two consecutive available seats without sitting next to Bob and Carla. So Alice has to sit in the other remaining chair. Then, there are two ways to seat Bob and Carla in the remaining two seats (which are consecutive). So this case gives us $8*2=16$ ways.

So in total there are $12+16=28$ . So our answer is $\boxed{\textbf{(C)}\ 28}$ .

Problem 20

Let $S(n)$ equal the sum of the digits of positive integer $n$ . For example, $S(1507) = 13$ . For a particular positive integer $n$ , $S(n) = 1274$ . Which of the following could be the value of $S(n+1)$ ?

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 1239\qquad\textbf{(E)}\ 1265$

Solution 1

Note that $n \equiv S(n) \pmod{9}$ . This can be seen from the fact that $\sum_{k=0}^{n}10^{k}a_k \equiv \sum_{k=0}^{n}a_k \pmod{9}$ . Thus, if $S(n) = 1274$ , then $n \equiv 5 \pmod{9}$ , and thus $n+1 \equiv S(n+1) \equiv 6 \pmod{9}$ . The only answer choice that is $6 \pmod{9}$ is $amc考试$ .

Solution 2

One divisibility rule for division that we can use in the problem is that a multiple of $9$ has its digit always add up to a multiple of $9$ . We can find out that the least number of digits the number $N$ is $142$ , with $141$ $9$ 's and $1$ $5$ , assuming the rule above. No matter what arrangement or different digits we use, the divisor rule stays the same. To make the problem simpler, we can just use the $141$ $9$ 's and $1$ $5$ . By randomly mixing the digits up, we are likely to get: $9999$ ... $9995999$ ... $9999$ . By adding $1$ to this number, we get: $9999$ ... $9996000$ ... $0000$ . Knowing that this number is ONLY divisible by $9$ when $6$ is subtracted, we can subtract $6$ from every available choice, and see if the number is divisible by $9$ afterwards. After subtracting $6$ from every number, we can conclude that $1233$ (originally $1239$ ) is the only number divisible by $9$ . So our answer is $\boxed{\textbf{(D)}\ 1239}$ .

Solution 3

The number $n$ can be viewed as having some unique digits in the front, following by a certain number of nines. We can then evaluate each potential answer choice.

If $\boxed{\textbf{(A)}\ 1}$ is correct, then $n$ must be some number $99999999...9$ , because when we add one to $99999999...9$ we get $10000000...00$ . Thus, if $1$ is the correct answer, then the equation $9x=1274$ must have an integer solution (i.e. $1274$ must be divisible by $9$ ). But since it does not, $1$ is not the correct answer.

If $\boxed{\textbf{(B)}\ 3}$ is correct, then $n$ must be some number $29999999...9$ , because when we add one to $29999999...9$ , we get $30000000...00$ . Thus, if $3$ is the correct answer, then the equation $2+9x=1274$ must have an integer solution. But since it does not, $3$ is not the correct answer.

Based on what we have done for evaluating the previous two answer choices, we can create an equation we can use to evaluate the final three possibilities. Notice that if $S(n+1)=N$ , then $n$ must be a number whose initial digits sum to $N-1$ , and whose other, terminating digits, are all $9$ . Thus, we can evaluate the three final possibilities by seeing if the equation $(N-1)+9x=1274$ has an integer solution.

The equation does not have an integer solution for $N=12$ , so $amc数学竞赛$ is not correct. However, the equation does have an integer solution for $N=1239$ ( $amc数学竞赛真题$ ), so $\boxed{\textbf{(D)}\ 1239}$ is the answer.