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AMC数学竞赛真题2016年B 5-6

2018-11-19 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 5

The mean age of Amanda's $4$ cousins is $8$ , and their median age is $5$ . What is the sum of the ages of Amanda's youngest and oldest cousins?

$AMC美国大学生数学竞赛$

Solution

The sum of the ages of the cousins is $4$ times the mean, or $32$ . There are an even number of cousins, so there is no single median, so $5$ must be the median of the two in the middle. Therefore the sum of the ages of the two in the middle is $10$ . Subtracting $10$ from $32$ produces $\textbf{(D)}\ 22$ .

Problem 6

Laura added two three-digit positive integers. All six digits in these numbers are different. Laura's sum is a three-digit number $S$ . What is the smallest possible value for the sum of the digits of $S$ ?

$AMC数学竞赛报考点$

Solution 1

Let the two three-digit numbers she added be $a$ and $b$ with $a+b=S$ and $a<b$ . The hundreds digits of these numbers must be at least $1$ and $2$ , so $a\ge 100$ and $b\ge 200$ .

Say $a=100+p$ and $b=200+q$ ; then we just need $p+q=100$ with $p$ and $q$ having different digits which aren't $1$ or $2$ .There are many solutions, but $p=3$ and $q=97$ give $103+297=400$ which proves that $\boxed{\textbf{(B)}\ 4}$ is attainable.

Solution 2

For this problem, to find the $3$ -digit integer with the smallest sum of digits, one should make the units and tens digit add to $0$ . To do that, we need to make sure the digits are all distinct. For the units digit, we can have a variety of digits that work. $7$ works best for the top number which makes the bottom digit $3$ . The tens digits need to add to $9$ because of the $1$ that needs to be carried from the addition of the units digits. We see that $5$ and $4$ work the best as we can't use $6$ and $3$ . Finally, we use $2$ and $1$ for our hundreds place digits.