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考题20-21 2015 AMC 12B

2018-08-06 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的官方真题以及官方解答吧：

Problem 20

For every positive integer $n$ , let $\text{mod}_5 (n)$ be the remainder obtained when $n$ is divided by 5. Define a function $f: \{0,1,2,3,\dots\} \times \{0,1,2,3,4\} \to \{0,1,2,3,4\}$ recursively as follows:

考题20-21 2015 AMC 12B

What is $f(2015,2)$ ?

$\textbf{(A)}\; 0 \qquad\textbf{(B)}\; 1 \qquad\textbf{(C)}\; 2 \qquad\textbf{(D)}\; 3 \qquad\textbf{(E)}\; 4$

Solution

Simply draw a table of values of $f(i,j)$ for the first few values of $i$ :

考题20-21 2015 AMC 12B

Now we claim that for $i \ge 5$ , $f(i,j) = 1$ for all values $0 \le j \le 4$ . We will prove this by induction on $i$ and $j$ . The base cases for $i = 5$ , have already been proven.

For our inductive step, we must show that for all valid values of $j$ , $f(i, j) = 1$ if for all valid values of $j$ , $f(i - 1, j) = 1$ .

We prove this itself by induction on $j$ . For the base case, $j=0$ , $f(i, 0) = f(i-1, 1) = 1$ . For the inductive step, we need $f(i, j) = 1$ if $f(i, j-1) = 1$ . Then, $f(i, j) = f(i-1, f(i, j-1)).$ $f(i, j-1) = 1$ by our inductive hypothesis from our inner induction and $f(i-1, 1) = 1$ from our outer inductive hypothesis. Thus, $f(i, j) = 1$ , completing the proof.

It is now clear that for $i \ge 5$ , $f(i,j) = 1$ for all values $0 \le j \le 4$ .

Thus, $f(2015,2) = \boxed{\textbf{(B)} \; 1}$ .

Problem 21

Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$ ?

$考题20-21 2015 AMC 12B$

Solution 1

We can translate this wordy problem into this simple equation:

$\left\lceil \frac{s}{2} \right\rceil - 19 = \left\lceil \frac{s}{5} \right\rceil$

We will proceed to solve this equation via casework.

Case 1: $\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}$

Our equation becomes $\frac{s}{2} - 19 = \frac{s}{5} + \frac{j}{5}$ , where $j \in \{0,1,2,3,4\}$ Using the fact that $s$ is an integer, we quickly find that $j=1$ and $j=4$ yield $s=64$ and $s=66$ , respectively.

Case 2: $考题20-21 2015 AMC 12B$

Our equation becomes $\frac{s}{2} +\frac{1}{2} - 19 = \frac{s}{5} + \frac{j}{5}$ , where $j \in \{0,1,2,3,4\}$ Using the fact that $s$ is an integer, we quickly find that $j=2$ yields $s=63$ .

Summing up we get $63+64+66=193$ . The sum of the digits is $\boxed{\textbf{(D)}\; 13}$ .

Solution 2

It can easily be seen that the problem can be expressed by the equation:

$\left\lceil \frac{s}{2} \right\rceil - \left\lceil \frac{s}{5} \right\rceil = 19$

However, because the ceiling function is difficult to work with, we can rewrite the previous equation as:

$\frac{s+a}{2} - \frac{s+b}{5} = 19$

Where $a \in \{0,1\}$ and $b \in \{0,1,2,3,4\}$ Multiplying both sides by ten and simplifying, we get:

$5s+5a-2s-2b=190$ $3s = 190+2b-5a$ $s = 63 + \frac{1+2b-5a}{3}$

Because s must be an integer, we need to find the values of $a$ and $b$ such that $2b-5a \equiv 2 \mod 3$ . We solve using casework.

Case 1: $a = 0$

If $a = 0$ , we have $2b \equiv 2 \mod 3$ . We can easily see that $b = 1$ or $b = 4$ , which when plugged into our original equation lead to $s = 64$ and $s=66$ respectively.

Case 2: $a = 1$

If $a = 1$ , we have $2b-5 \equiv 2 \mod 3$ , which can be rewritten as $2b \equiv 1 \mod 3$ . We can again easily see that $b = 2$ is the only solution, which when plugged into our original equation lead to $s = 63$ .

Adding these together we get $64+66+63=193$ . The sum of the digits is $\boxed{\textbf{(D)}\; 13}$ .

Solution 3

As before, we write the equation:

$\left\lceil \frac{s}{2} \right\rceil - 19 = \left\lceil \frac{s}{5} \right\rceil.$

To get a ballpark estimate of where $s$ might lie, we remove the ceiling functions to find:

$\frac{s}{2} - 19 = \frac{s}{5}.$

This gives $\frac{3s}{10} = 19$ , and thus values for $s$ will be around $\frac{190}{3} = 63.\overline3$ .

Now, to establish some bounds around this estimated working value, we note that if $s=60$ , Cozy takes 30 steps while Dash takes 12, a difference of 18. If $s=70$ , Cozy takes 35 steps while Dash takes 14, a difference of 21. When $s$ increases from a multiple of ten, the difference will never decrease beyond what it is at the multiple of ten, and likewise, when it decreases, it never becomes greater than at the multiple of ten, so any working values of $s$ will be between $60$ and $70$ .