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AMC数学竞赛真题2016年12A 19-20

2018-09-01 重点归纳

AMC12是针对高中学生的数学测验，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。其主要目的在于激发学生对数学的兴趣，参予AMC12的学生应该不难发现测验的问题都很具挑战性，但测验的题型都不会超过学生的学习范围。这项测验希望每个考生能从竞赛中享受数学。那么接下来跟随小编来看一下AMC12的官方真题以及官方解答吧：

Problem 19

Jerry starts at $0$ on the real number line. He tosses a fair coin $8$ times. When he gets heads, he moves $1$ unit in the positive direction; when he gets tails, he moves $1$ unit in the negative direction. The probability that he reaches $4$ at some time during this process $\frac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. What is $a + b?$ (For example, he succeeds if his sequence of tosses is $HTHHHHHH.$ )

$\textbf{(A)}\ 69\qquad\textbf{(B)}\ 151\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 293\qquad\textbf{(E)}\ 313$

Solution 1

For $6$ to $8$ heads, we are guaranteed to hit $4$ heads, so the sum here is $amc12$ .

For $4$ heads, you have to hit the $4$ heads at the start so there's only one way, $1$ .

For $5$ heads, we either start off with $4$ heads, which gives us $4\textbf{C}1=4$ ways to arrange the other flips, or we start off with five heads and one tail, which has $6$ ways minus the $2$ overlapping cases, $\text{HHHHHTTT}$ and $\text{HHHHTHTT}$ . Total ways: $8$ .

Then we sum to get $46$ . There are a total of $2^8=256$ possible sequences of $8$ coin flips, so the probability is $\frac{46}{256}=\frac{23}{128}$ . Summing, we get $23+128=\boxed{\textbf{(B) }151}$ .

Solution 2

Reaching 4 will require either 4, 6, or 8 flips. Therefore we can split into 3 cases:

(Case 1): The first four flips are heads. Then, the last four flips can be anything so $2^4=16$ possibilities work.

(Case 2): It takes 6 flips to reach 4. There must be one tail in the first four flips so we don't repeat case 1. The tail can be in one of 4 positions. The next two flips must be heads. The last two flips can be anything so $2^2=4$ flips work. $4*4=16$ .

(Case 3): It takes 8 flips to reach 4. We can split this case into 2 sub-cases. There can either be 1 or 2 tails in the first 4 flips.

(1 tail in first four flips). In this case, the first tail can be in 4 positions. The second tail can be in either the 5th or 6th position so we don't repeat case 2. Thus, there are $4*2=8$ possibilities.

(2 tails in first four flips). In this case, the tails can be in $\binom{4}{2}=6$ positions.

Adding these cases up and taking the total out of $2^8=256$ yields $美国数学竞赛amc12真题$ . This means the answer is $23+128=\boxed{\textbf{(B) }151}$ .

Problem 20

A binary operation $\diamondsuit$ has the properties that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = (a\,\diamondsuit \,b)\cdot c$ and that $a\,\diamondsuit \,a=1$ for all nonzero real numbers $a, b,$ and $c$ . (Here $\cdot$ represents multiplication). The solution to the equation $2016 \,\diamondsuit\, (6\,\diamondsuit\, x)=100$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. What is $p+q?$

$\textbf{(A) }109\qquad\textbf{(B) }201\qquad\textbf{(C) }301\qquad\textbf{(D) }3049\qquad\textbf{(E) }33,601$

Solutions

Solution 1

We see that $a \diamond a = 1$ , and think of division. Testing, we see that the first condition $a \diamond (b \diamond c) = (a \diamond b) \cdot c$ is satisfied, because $\frac{a}{\frac{b}{c}} = \frac{a}{b} \cdot c$ . Therefore, division is the operation $\diamond$ . Solving the equation, $\frac{2016}{\frac{6}{x}} = \frac{2016}{6} \cdot x = 336x = 100\implies x=\frac{100}{336} = \frac{25}{84},$ so the answer is $25 + 84 = \boxed{\textbf{(A) }109}$ .

Solution 2

We can manipulate the given identities to arrive at a conclusion about the binary operator $\diamondsuit$ . Substituting $b = c$ into the first identity yields $amc数学竞赛$ Hence, $( a\ \diamondsuit\ b) \cdot b = a,$ or, dividing both sides of the equation by $b,$ $( a\ \diamondsuit\ b) = \frac{a}{b}.$

Hence, the given equation becomes $\frac{2016}{\frac{6}{x}} = 100$ . Solving yields $x=\frac{100}{336} = \frac{25}{84},$ so the answer is $25 + 84 = \boxed{\textbf{(A) }109.}$

Solution 3

One way to eliminate the $\diamondsuit$ in this equation is to make $a = b$ so that $a\,\diamondsuit\, (b\,\diamondsuit \,c) = c$ . In this case, we can make $b = 2016$ .