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考题23-25 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 23

Let $n$ be a positive integer greater than 4 such that the decimal representation of $n!$ ends in $k$ zeros and the decimal representation of $(2n)!$ ends in $3k$ zeros. Let $s$ denote the sum of the four least possible values of $n$ . What is the sum of the digits of $s$ ?

$考题4-6 2015 AMC 10B$

Solution 1

A trailing zero requires a factor of two and a factor of five. Since factors of two occur more often than factors of five, we can focus on the factors of five. We make a chart of how many trailing zeros factorials have:

$考题4-6 2015 AMC 10B$

We first look at the case when $n!$ has $1$ zero and $(2n)!$ has $3$ zeros. If $n=5,6,7$ , $(2n)!$ has only $2$ zeros. But for $n=8,9$ , $(2n)!$ has $3$ zeros. Thus, $n=8$ and $n=9$ work.

Secondly, we look at the case when $n!$ has $2$ zeros and $(2n)!$ has $6$ zeros. If $n=10,11,12$ , $(2n)!$ has only $4$ zeros. But for $n=13,14$ , $(2n)!$ has $6$ zeros. Thus, the smallest four values of $n$ that work are $n=8,9,13,14$ , which sum to $44$ . The sum of the digits of $44$ is $\boxed{\mathbf{(B)\ }8}$

Solution 2

By Legendre's Formula and the information given, we have that $考题4-6 2015 AMC 10B$ .

Trivially, it is obvious that $n<100$ as there is no way that if $n>100$ , $n!$ would have $3$ times as many zeroes as $(2n)!$ .

First, let's plug in the number $5$ We get that $3(1)=1$ , which is obviously not true. Hence, $n>5$

After several attempts, we realize that the RHS needs $1$ to $2$ more "extra" zeroes than the LHS. Hence, $n$ is greater than a multiple of $5$ .

Very quickly, we find that the least $4 n's$ are $8,9,13,14$ .

$考题4-6 2015 AMC 10B$ .

Problem 24

Aaron the ant walks on the coordinate plane according to the following rules. He starts at the origin $p_0=(0,0)$ facing to the east and walks one unit, arriving at $p_1=(1,0)$ . For $n=1,2,3,\dots$ , right after arriving at the point $p_n$ , if Aaron can turn $90^\circ$ left and walk one unit to an unvisited point $p_{n+1}$ , he does that. Otherwise, he walks one unit straight ahead to reach $p_{n+1}$ . Thus the sequence of points continues $p_2=(1,1), p_3=(0,1), p_4=(-1,1), p_5=(-1,0)$ , and so on in a counterclockwise spiral pattern. What is $p_{2015}$ ?

$考题4-6 2015 AMC 10B$

Solution 1

The first thing we would do is track Aaron's footsteps:

He starts by taking $1$ step East and $1$ step North, ending at $(1,1)$ after $2$ steps and about to head West.

Then he takes $2$ steps West and $2$ steps South, ending at $(-1,-1$ ) after $2+4$ steps, and about to head East.

Then he takes $3$ steps East and $3$ steps North, ending at $(2,2)$ after $2+4+6$ steps, and about to head West.

Then he takes $4$ steps West and $4$ steps South, ending at $(-2,-2)$ after $2+4+6+8$ steps, and about to head East.

From this pattern, we can notice that for any integer $k \ge 1$ he's at $(-k, -k)$ after $2 + 4 + 6 + ... + 4k$ steps, and about to head East. There are $2k$ terms in the sum, with an average value of $(2 + 4k)/2 = 2k + 1$ , so:

$2 + 4 + 6 + ... + 4k = 2k(2k + 1)$

If we substitute $k = 22$ into the equation: $44(45) = 1980 < 2015$ . So he has $35$ moves to go. This makes him end up at $考题4-6 2015 AMC 10B$ .

Solution 2

We are given that Aaron starts at $(0, 0)$ , and we note that his net steps follow the pattern of $+1$ in the $x$ -direction, $+1$ in the $y$ -direction, $-2$ in the $x$ -direction, $-2$ in the $y$ -direction, $+3$ in the $x$ -direction, $+3$ in the $y$ -direction, and so on, where we add odd and subtract even.

We want $2 + 4 + 6 + 8 + ... + 2n = 2015$ , but it does not work out cleanly. Instead, we get that $2 + 4 + 6 + ... + 2(44) = 1980$ , which means that there are $35$ extra steps past adding $-44$ in the $x$ -direction (and the final number we add in the $y$ -direction is $-44$ ).

So $p_{2015} = (0+1-2+3-4+5...-44+35, 0+1-2+3-4+5...-44)$ .

We can group $1-2+3-4+5...-44$ as $考题4-6 2015 AMC 10B$ .

Thus $p_{2015} = \boxed{\textbf{(D)}\; (13, -22)}$ .

Solution 3

Looking at his steps, we see that he walks in a spiral shape. At the $8$ th step, he is on the bottom right corner of the $3\times 3$ square centered on the origin. On the $24$ th step, he is on the bottom right corner of the $5\times 5$ square centered at the origin. It seems that the $p_{n^2-1}$ is the bottom right corner of the $n\times n$ square. This makes sense since, after $n^2-1$ , he has been on n^2 dots, including the point $p_{0}$ . Also, this is only for odd $n$ , because starting with the $1\times 1$ square, we can only add one extra set of dots to each side, so we cannot get even $n$ . Since $45^2=2025$ , $p_{2024}$ is the bottom right corner of the $45\times 45$ square. This point is $\frac{45-1}{2}=22$ over to the right, and therefore $22$ down, so $p_{2024}=(22, -22)$ . Since $p_{2024}$ is $9$ ahead of $p_{2015}$ , we go back $9$ spaces to $\boxed{\textbf{(D)}\; (13, -22)}$ .

Problem 25

A rectangular box measures $a \times b \times c$ , where $a$ , $b$ , and $c$ are integers and $1\leq a \leq b \leq c$ . The volume and the surface area of the box are numerically equal. How many ordered triples $(a,b,c)$ are possible?

$考题4-6 2015 AMC 10B$

Solution

The surface area is $2(ab+bc+ca)$ , the volume is $abc$ , so $2(ab+bc+ca)=abc$ .

Divide both sides by $2abc$ , we have:

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2}.$

First consider the bound of the variable $a$ . Since $\frac{1}{a}<\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2},$ we have $a>2$ , or $a\ge 3$ .

Also note that $c\ge b\ge a>0$ , we have $\frac{1}{a}>\frac{1}{b}>\frac{1}{c}$ . Thus, $考题4-6 2015 AMC 10B$ , so $a\le 6$ .

So we have $a=3, 4, 5$ or $6$ .

Before the casework, let's consider the possible range for $b$ if $\frac{1}{b}+\frac{1}{c}=k>0$ .

From $\frac{1}{b}<k$ , we have $b>\frac{1}{k}$ . From $\frac{2}{b}\ge \frac{1}{b}+\frac{1}{c}=k$ , we have $b\le \frac{2}{k}$ . Thus $\frac{1}{k}<b\le \frac{2}{k}$

When $a=3$ , $\frac{1}{b}+\frac{1}{c}=\frac{1}{6}$ , so $b=7, 8, \cdots, 12$ . The solutions we find are $(a, b, c)=(3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), (3, 12, 12)$ , for a total of $5$ solutions.

When $a=4$ , $\frac{1}{b}+\frac{1}{c}=\frac{1}{4}$ , so $b=5, 6, 7, 8$ . The solutions we find are $(a, b, c)=(4, 5, 20), (4, 6, 12), (4, 8, 8)$ , for a total of $3$ solutions.

When $a=5$ , $\frac{1}{b}+\frac{1}{c}=\frac{3}{10}$ , so $b=5, 6$ . The only solution in this case is $(a, b, c)=(5, 5, 10)$ .

When $a=6$ , $b$ is forced to be $6$ , and thus $(a, b, c)=(6, 6, 6)$ .

Thus, our answer is $\boxed{\textbf{(B)}\;10}$

Simplification of Solution

Start right after the Solution says $a$ is $3, 4, 5, 6$ . We can say $\frac{1}{b}+\frac{1}{c}=\frac{1}{q}$ , where $考题4-6 2015 AMC 10B$ .

Notice $immediately$ that $b, c > q$ ! This is our key step. Then we can say $b=q+d$ , $c=q+e$ . If we clear the fraction about b and c (do the math), our immediate result is that $de = q^2$ . Realize also that $d<e$ .

Now go through cases for $a$ and you end up with the same result. However, now you don't have to guess solutions. For example, when $a=3$ , then $de = 36$ and $考题4-6 2015 AMC 10B$ .