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AMC数学竞赛真题2016年12B 15-16

2018-09-10 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AMC真题以及官方解答吧：

AMC真题 15

All the numbers $2, 3, 4, 5, 6, 7$ are assigned to the six faces of a cube, one number to each face. For each of the eight vertices of the cube, a product of three numbers is computed, where the three numbers are the numbers assigned to the three faces that include that vertex. What is the greatest possible value of the sum of these eight products?

$amc真题$

AMC真题解答 1

First assign each face the letters $a,b,c,d,e,f$ . The sum of the product of the faces is $abc+acd+ade+aeb+fbc+fcd+fde+feb$ . We can factor this into $(a+f)(b+c)(d+e)$ which is the product of the sum of each pair of opposite faces. In order to maximize $(a+f)(b+c)(d+e)$ we use the numbers $amc真题$ or $amc真题$ .

AMC真题解答 2

We proceed from the factorization in the above solution. By the AM-GM inequality,

$\frac{a_1+a_2+a_3}{3}\geq\sqrt[3]{a_1a_2a_3}$

Cubing both sides,

$\left(\frac{a_1+a_2+a_3}{3}\right)^3\geq{a_1a_2a_3}$

Let $a_1=(a+f)$ , $a_2=(b+c)$ , and $a_3=(d+e)$ . Let's substitute in these values.

$amc真题$

$a+b+c+d+e+f$ is fixed at 27.

$amc真题$

$\boxed{\textbf{(D)}\ 729 }\geq{(a+f)(b+c)(d+e)}$

AMC真题 16

In how many ways can $345$ be written as the sum of an increasing sequence of two or more consecutive positive integers?

$amc真题$

AMC真题解答1

We proceed with this problem by considering two cases, when: 1) There are an odd number of consecutive numbers, 2) There are an even number of consecutive numbers.

For the first case, we can cleverly choose the convenient form of our sequence to be $a-n,\cdots, a-1, a, a+1, \cdots, a+n$

because then our sum will just be $(2n+1)a$ . We now have $(2n+1)a = 345$ and $a$ will have a solution when $\frac{345}{2n+1}$ is an integer, namely when $2n+1$ is a divisor of 345. We check that $2n+1 = 3, 5, 15, 23$ work, and no more, because $2n+1=1$ does not satisfy the requirements of two or more consecutive integers, and when $2n+1$ equals the next biggest factor, $69$ , there must be negative integers in the sequence. Our solutions are $amc真题$ .

For the even cases, we choose our sequence to be of the form: $a-(n-1), \cdots, a, a+1, \cdots, a+n$ so the sum is $\frac{(2n)(2a+1)}{2} = n(2a+1)$ . In this case, we find our solutions to be $\{172, 173\}, \{55,\cdots, 60\}, \{30,\cdots, 39\}$ .

We have found all 7 solutions and our answer is $amc真题$ .

AMC真题解答2

The sum from $a$ to $b$ where $a$ and $b$ are integers and $a>b$ is

$S=\dfrac{(a-b+1)(a+b)}{2}$

$345=\dfrac{(a-b+1)(a+b)}{2}$

$2\cdot 3\cdot 5\cdot 23=(a-b+1)(a+b)$

Let $c=a-b+1$ and $d=a+b$

$2\cdot 3\cdot 5\cdot 23=c\cdot d$

If we factor $690$ into all of its factor groups $(\text{exg}~ (10,69) ~\text{or} ~(15,46))$ we will have several ordered pairs $(c,d)$ where $c<d$

The number of possible values for $c$ is half the number of factors of $690$ which is $\frac{1}{2}\cdot2\cdot2\cdot2\cdot2=8$

However, we have one extraneous case of $(1,690)$ because here, $a=b$ and we have the sum of one consecutive number which is not allowed by the question.

Thus the answer is $amc真题$

$\boxed{\textbf{(E)} \,7}$ .

AMC真题解答3

There is a handy formula for this problem: The number of odd factors of $345$

$345 = 5*3*23$

$2^3 = 8$

There are 8 ways to have an increasing sum of positive integers that add to 345. However, we have to subtract one for the case where it is just $345$ . The problem wants two or more consecutive integers.

Therefore, $8-1=$ $\boxed{\textbf{(E)} \,7}$ .

AMC真题解答4

We're dealing with an increasing arithmetic progression of common difference 1. Let $x$ be the number of terms in a summation. Let $y$ be the first term in a summation. The sum of an arithmetic progression is the average of the first term and the last term multiplied by the number of terms. The problem tells us that the sum must be 345.

$amc真题$

In order to satisfy the constraints of the problem, x and y must be positive integers. Maybe we can make this into a Diophantine thing! In fact, if we just factor out that $x$ ... voilà!

$(x)(x+2y-1)=690$

There are 16 possible factor pairs to try (for brevity, I will not enumerate them here). Notice that the expression in the right parenthesis is $2y-1$ more than the expression in the parenthesis on the left. $y$ is at least 1. Thus, the expression in the right parenthesis will always be greater than the expression on the left. This eliminates 8 factor pairs. The problem also says the "increasing sequence" has to have "two or more" terms, so $x \geqslant 2$ . This eliminates the factor pair $1 \cdot 690$ . With brief testing, we find that the the other 7 factor pairs produce 7 viable ordered pairs. This means we have found $\boxed{\textbf{(E)}\ 7}$ ways to write 345 in the silly way outlined by the problem.