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AMC数学竞赛真题2016年B 18

2018-11-30 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 18

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

$AMC数学竞赛$

Solution 1

Factor $345=3\cdot 5\cdot 23$ .

Suppose we take an odd number $k$ of consecutive integers, with the median as $m$ . Then $mk=345$ with $\tfrac12k<m$ . Looking at the factors of $345$ , the possible values of $k$ are $3,5,15,23$ with medians as $115,69,23,15$ respectively.

Suppose instead we take an even number $2k$ of consecutive integers, with median being the average of $m$ and $m+1$ . Then $k(2m+1)=345$ with $k\le m$ . Looking again at the factors of $345$ , the possible values of $k$ are $1,3,5$ with medians $(172,173),(57,58),(34,35)$ respectively.

Thus the answer is $\textbf{(E) }7$ .

Solution 2

We need to find consecutive numbers (an arithmetic sequence that increases by $1$ ) that sums to $345$ . This calls for the sum of an arithmetic sequence given that the first term is $k$ , the last term is $g$ and with $n$ elements, which is: $美国数学竞赛amc网站$ .

So, since it is a sequence of $n$ consecutive numbers starting at $k$ and ending at $k+n-1$ . We can now substitute $g$ with $k+n-1$ . Now we substiute our new value of $g$ into $\frac {n*(k+g)}{2}$ to get that the sum is $\frac {n*(k+k+n-1)}{2} = 345$ .

This simplifies to $\frac {n*(2k+n-1)}{2} = 345$ . This gives a nice equation. We multiply out the 2 to get that $n*(2k+n-1)=690$ . This leaves us with 2 integers that multiplies to $690$ which leads us to think of factors of $690$ . We know the factors of $690$ are: $1,2,3,5,6,10,15,23,30,46,69,115,138,230,345,690$ . So through inspection (checking), we see that only $2,3,5,6,10,15$ and $23$ work. This gives us the answer of $\textbf{(E) }7$ ways.