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考题11-12 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 11

Among the positive integers less than 100, each of whose digits is a prime number, one is selected at random. What is the probability that the selected number is prime?

$考题11-12 2015 AMC 10B$

Solution

The one digit prime numbers are $2$ , $3$ , $5$ , and $7$ . So there are a total of $4\cdot4=16$ ways to choose a two digit number with both digits as primes and 4 ways to choose a one digit prime, for a total of $4+16=20$ ways. Out of these $2$ , $3$ , $5$ , $7$ , $23$ , $37$ , $53$ , and $73$ are prime. Thus the probability is $\dfrac{8}{20}=\boxed{\textbf{(B)} \dfrac{2}{5}}$ .

Problem 12

For how many integers $x$ is the point $(x, -x)$ inside or on the circle of radius 10 centered at $(5, 5)$ ?

$考题11-12 2015 AMC 10B$

Solution

The equation of the circle is $(x-5)^2+(y-5)^2=100$ . Plugging in the given conditions we have $(x-5)^2+(-x-5)^2 \leq 100$ . Expanding gives: $x^2-10x+25+x^2+10x+25\leq 100$ , which simplifies to $x^2\leq 25$ and therefore $x\leq 5$ or $x\geq -5$ . So $x$ ranges from $-5$ to $5$ , for a total of $\boxed{\mathbf{(A)}\ 11}$ values.