首页> 重点归纳 > 考题17-18 2015 AMC 12B

考题17-18 2015 AMC 12B

2018-08-06 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 17

An unfair coin lands on heads with a probability of $\tfrac{1}{4}$ . When tossed $n>1$ times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the value of $n$ ?

$考题17-18 2015 AMC 12B$

Solution

When tossed $n$ times, the probability of getting exactly 2 heads and the rest tails is

考题17-18 2015 AMC 12B

Similarly, the probability of getting exactly 3 heads is

$\dbinom{n}{3}{\left( \frac{1}{4} \right)}^3 {\left( \frac{3}{4} \right) }^{n-3}.$

Now set the two probabilities equal to each other and solve for $n$ :

考题17-18 2015 AMC 12B

$3 = \frac{n-2}{3}$

$n-2 = 9$

$n = \fbox{\textbf{(D)}\; 11}$

Note: the original problem did not specify $n>1$ , so $n=1$ was a solution, but this was fixed in the Wiki problem text so that the answer would make sense.

Problem 18

For every composite positive integer $n$ , define $r(n)$ to be the sum of the factors in the prime factorization of $n$ . For example, $r(50) = 12$ because the prime factorization of $50$ is $2 \times 5^{2}$ , and $2 + 5 + 5 = 12$ . What is the range of the function $r$ , $\{r(n): n \text{ is a composite positive integer}\}$ ?

$考题17-18 2015 AMC 12B$

Solution

Solution 1

This problem becomes simple once we recognize that the domain of the function is $\{4, 6, 8, 9, 10, 12, 14, 15, \dots\}$ . By evaluating $r(4)$ to be $4$ , we can see that $\textbf{(E)}$ is incorrect. Evaluating $r(6)$ to be $5$ , we see that both $\textbf{(B)}$ and $\textbf{(C)}$ are incorrect. Since our domain consists of composite numbers, which, by definition, are a product of at least two positive primes, the minimum value of $r(n)$ is $4$ , so $\textbf{(A)}$ is incorrect. That leaves us with $\boxed{\textbf{(D)}\; \text{the set of integers greater than }3}$ .

Solution 2

Think backwards. The range is the same as the numbers $y$ that can be expressed as the sum of two or more prime positive integers.

The lowest number we can get is $y = 2+2 = 4$ . For any number greater than 4, we can get to it by adding some amount of 2's and then possibly a 3 if that number is odd. For example, 23 can be obtained by adding 2 ten times and adding a 3; this corresponds to the argument $n = 2^{10} \times 3$ . Thus our answer is $\boxed{\textbf{(D)}\; \text{the set of integers greater than }3}$ .