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AMC数学竞赛真题2017年10B 13-14

2018-08-30 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 13

There are $20$ students participating in an after-school program offering classes in yoga, bridge, and painting. Each student must take at least one of these three classes, but may take two or all three. There are $10$ students taking yoga, $13$ taking bridge, and $9$ taking painting. There are $9$ students taking at least two classes. How many students are taking all three classes?

$amc真题$

Solution

By PIE (Property of Inclusion/Exclusion), the answer is $amc数学竞赛$ .

Problem 14

An integer $N$ is selected at random in the range $1\leq N \leq 2020$ . What is the probability that the remainder when $N^{16}$ is divided by $5$ is $1$ ?

$美国数学竞赛$

Solution 1

By Fermat's Little Theorem, $N^{16} = (N^4)^4 \equiv 1 \text{ (mod 5)}$ when N is relatively prime to 5. However, this happens with probability $amc美国数学竞赛$ .

Solution 2

Note that the patterns for the units digits repeat, so in a sense we only need to find the patterns for the digits $0-9$ . The pattern for $0$ is $0$ , no matter what power, so $0$ doesn't work. Likewise, the pattern for $5$ is always $5$ . Doing the same for the rest of the digits, we find that the units digits of $1^{16}$ , $2^{16}$ , $3^{16}$ , $4^{16}$ , $6^{16}$ , $7^{16}$ , $8^{16}$ and $9^{16}$ all have the remainder of $1$ when divided by $5$ , so $\boxed{\textbf{(D) } \frac 45}$ .

Solution 3 (Casework)

We can use modular arithmetic for each residue of $n \pmod 5$

If $n \equiv 0 \pmod 5$ , then $n^{16} \equiv 0^{16} \equiv 0 \pmod 5$

If $n \equiv 1 \pmod 5$ , then $amc真题$

If $n \equiv 2 \pmod 5$ , then $n^{16} \equiv (n^2)^8 \equiv (2^2)^8 \equiv 4^8 \equiv (-1)^8 \equiv 1 \pmod 5$

If $n \equiv 3 \pmod 5$ , then $n^{16} \equiv (n^2)^8 \equiv (3^2)^8 \equiv 9^8 \equiv (-1)^8 \equiv 1 \pmod 5$

If $n \equiv 4 \pmod 5$ , then $n^{16} \equiv 4^{16} \equiv (-1)^{16} \equiv 1 \pmod 5$

In $4$ out of the $5$ cases, the result was $1 \pmod 5$ , and since each case occurs equally as $2020 \equiv 0 \pmod 5$ , the answer is $\boxed{\textbf{(D) }\frac{4}{5}}$