AMC10 2023 B

AMC10 2023 B · Q16

AMC10 2023 B · Q16. It mainly tests Basic counting (rules of product/sum), Combinations.

Define an $\textit{upno}$ to be a positive integer of $2$ or more digits where the digits are strictly increasing moving left to right. Similarly, define a $\textit{downno}$ to be a positive integer of $2$ or more digits where the digits are strictly decreasing moving left to right. For instance, the number $258$ is an upno and $8620$ is a downno. Let $U$ equal the total number of $upnos$ and let $D$ equal the total number of $downnos$. What is $|U-D|$?

将一个正整数定义为\textit{upno}，如果它有$2$位或更多位，且从左到右数字严格递增。类似地，将一个正整数定义为\textit{downno}，如果它有$2$位或更多位，且从左到右数字严格递减。例如，数字$258$是一个upno，$8620$是一个downno。令$U$为所有upno的总数，$D$为所有downno的总数。求$|U-D|$？

(A) 512 512

(B) 10 10

(D) 9 9

(E) 511 511

Answer

Correct choice: (E)

正确答案：（E）

Solution

First, we know that $D$ is greater than $U$, since there are less $upnos$ than $downnos$. To see why, we examine what determines an upno or $downno$. We notice that, given any selection of unique digits (notice that "unique" constrains this to be a finite number), we can construct a unique downno. Similarly, we can also construct an $upno$, but the selection can not include the digit $0$ since that isn't valid. We then have $[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]$ as the pool of numbers that we can pick from for $downnos$, and {1, 2, 3, 4, 5, 6, 7, 8, 9} as the pool of numbers that we can pick for $upnos$. There are only two states each number can be in: appearing and not appearing in the arrangement. (That is why we use the number 2 as the base in the exponent!) Thus, there are $2^{10}$ total $downnos$ and $2^9$ total $upnos$. However, we are told that each $upno$ or $downno$ must be at least $2$ digits, so we subtract out the $0$-digit and $1$-digit cases (Referring back to Paragraph 2, this is when every number's state is nonappearing or every number except one has the state of nonappearing). For the $downnos$, there are $10$ $1$-digit cases, and for the $upnos$, there are $9$ $1$-digit cases. There is $1$ $0$-digit case for both $upnos$ and $downnos$. Thus, the difference is $\left(\left(2^{10}-10-1\right)-\left(2^9-9-1\right)\right)=2^9-1=\boxed{\textbf{(E) }511}.$

首先，我们知道$D>U$，因为upno比downno少。要理解原因，我们考察决定upno或downno的因素。我们注意到，对于任何选择的唯一数字（注意“唯一”限制了有限数量），我们可以构造一个唯一的downno。类似地，我们也可以构造一个upno，但选择不能包括数字$0$，因为那无效。因此，对于downno，我们有$[0,1,2,3,4,5,6,7,8,9]$作为可选数字池，对于upno是$\{1,2,3,4,5,6,7,8,9\}$。每个数字只有两种状态：出现或不出现。（这就是为什么使用$2$作为指数的底数！）因此，downno总数为$2^{10}$，upno总数为$2^9$。但是，upno或downno必须至少$2$位，所以我们减去$0$位和$1$位的情况（回顾第二段，这是每个数字都不出现或除了一个外都不出现的情况）。对于downno，有$10$个$1$位情况，对于upno有$9$个$1$位情况。两者都有$1$个$0$位情况。因此，差值为$\left((2^{10}-10-1)-(2^9-9-1)\right)=2^9-1=\boxed{\textbf{(E) }511}$。

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