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

Revision as of 20:11, 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 1

$D$ 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 $x = \sum_{k=1}^9 {9 \choose k} = \sum_{k=0}^9 {9 \choose k} - {9 \choose 0} = 2^9-1 = \textbf{511}$ .

~Technodoggo ~minor edits by lucaswujc

Solution 2

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 3

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)

Solution 4 (Educated Guess)

First, note that the only $downnos$ that are not contained by the set of $upnos$ is every $downno$ that ends in $0$ .

Next, listing all the two digits $downnos$ , we find that the answer is more than 9, since there are more digits to be tested and there are 9 two digit $downnos$ . This leaves us with $512$ or $511$ .

Next, we notice that all the possibilities for $2$ through $9$ digit $downnos$ ending in $0$ pair up with one another, as the possibilities are equal (e.g. possibilities for $2$ digits = possibilities for $9$ digits, etc.).

This leaves us with one last possibility, the ten digit $downno 9876543210$ .

Since all the previous possibilities form an even number, adding one more possibility will make the total odd. Therefore, we need to choose the odd number from the set $[511, 512]$ .

Our answer is 511.

~yourmomisalosinggame (a.k.a. Aaron)

@@ Line 7: / Line 7: @@
 == Solution 1 ==
 <math>D</math> is greater than <math>U</math> because <math>upno</math> can't start with <math>0</math>.
-So the differences are in the form <math>x0</math>
+So the differences are in the form <math>x0</math>.
 When <math>x</math> has length <math>k</math> we have  <math>{9 \choose k}.</math>
 The number of possible <math>x = \sum_{k=1}^9 {9 \choose k} = \sum_{k=0}^9 {9 \choose k} - {9 \choose 0} = 2^9-1 = \textbf{511}</math>.

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