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Difference between revisions of "2023 AMC 10B Problems/Problem 16"

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== Solution 2 ==
 
== Solution 2 ==
  
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Note that you can obtain a downo by reversing an upno (like 235 is an upno, and you can obtain 532). So, we need to find the amount of downos that end with 0. We can use stars and bars to get 9 choose 0 + 9 choose 1 + 9 choose 2... + 9 choose 9 to get 512. However, 0 is not a valid case so we subtract 1. Our answer is 511.
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Note that you can obtain a downo by reversing an upno (like 235 is an upno, and you can obtain 532). So, we need to find the amount of downos that end with 0. We can use stars and bars to get: <cmath>\sum_{n=0}^9 \dbinom{9}{n} = \dbinom{9}{0} + \dbinom{9}{1} + \cdots + \dbinom{9}{9}</cmath> to get 512. However, 0 is not a valid case so we subtract 1. Our answer is 511.
  
 
-aleyang
 
-aleyang
  
−
(Sorry this solution is really bad, I am not that good at latex and this is my first time doing this.)
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-ap246(LaTeX)

Revision as of 18:21, 15 November 2023

Problem

Define an $upno$ to be a positive integer of 2 or more digits where the digits are strictly increasing moving left to right. Similarly, define a $downno$ 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|?

Solution

$D$ is greater than $U$ because $upno$ can't start with $0$. So the differences are in the form $x0$ When x has length $k$ we have ${9 \choose k}.$ The number of possible $x = \sum_{k=1}^9 {9 \choose k} = \sum_{k=0}^9 {9 \choose k} - {9 \choose 0} = 2^9-1 = 511$.

~Technodoggo

Solution

Since Upnos do not allow 0s to be in their first -- and any other -- digit, there will be no zeros in any digits of an Upno. Thus, Upnos only contain digits [1,2,3,4,5,6,7,8,9].

Upnos are 2 digits in minimum and 9 digits maximum (repetition is not allowed). Thus the total number of Upnos will be (9C2)+(9C3)+(9C4)+...+(9C9), since every selection of distinct numbers from the set [1,2,3,4,5,6,7,8,9] can be arranged so that it is an Upno. There will be (9C2) 2 digit Upnos, (9C3) 3 digit Upnos and so on.

Thus, the total number of Upnos will be (9C2)+(9C3)+(9C4)+...+(9C9) = 2^9-(9C0)-(9C1) = 512 - 10 = 502.

Notice that the same combination logic can be done for Downnos, but Downnos DO allow zeros to be in their last digit. Thus, there are 10 possible digits [0,1,2,3,4,5,6,7,8,9] for Downnos.

Therefore, it is visible that the total number of Downnos are (10C2)+(10C3)+(10C4)+...+(10C10) = 2^10-(10C0)-(10C10) = 1024 - 11 = 1013.

Thus abs(#Upno-#Downno) = abs(1013-502) = 511.


~yxyxyxcxcxcx


Solution 2

Note that you can obtain a downo by reversing an upno (like 235 is an upno, and you can obtain 532). So, we need to find the amount of downos that end with 0. We can use stars and bars to get: \[\sum_{n=0}^9 \dbinom{9}{n} = \dbinom{9}{0} + \dbinom{9}{1} + \cdots + \dbinom{9}{9}\] to get 512. However, 0 is not a valid case so we subtract 1. Our answer is 511.

-aleyang

-ap246(LaTeX)

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