2023 AMC 10B Problems/Problem 16


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 $|U-D|$?


Video Solution by MegaMath


Solution 1

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.

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.

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 $\left(\left(2^{10}-10-1\right)-\left(2^9-9-1\right)\right)=2^9-1=\boxed{\textbf{(E) }511}.$

~Technodoggo ~minor edits by lucaswujc

Solution 2

Since Upnos do not allow $0$s 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 ${9\choose 2}+{9\choose 3}+ {9\choose 4}+...+{9\choose 9}$, 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 ${9\choose 2}$ -$2$ digit Upnos, ${9\choose 3}$ -$3$ digit Upnos and so on.

Thus, the total number of Upnos will be $\sum_{i=2}^{9}{{9 \choose i}}=512-10=502$

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: \[\sum_{i=2}^{10}{{10\choose i}}= 2^{10}-{10\choose 0}-{10\choose 1} = 1024 - 11 = 1013\].

Thus $|U-D| = |(1013-502)| =\boxed{\textbf{(E) }511}.$

~yxyxyxcxcxcx ~JISHNU4414L (Latex)

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 since if you 'flip' the numbers, the upno starts with a 0 which corresponds to a downo. We can find the cases that end with a 0: \[\sum_{n=0}^9 \dbinom{9}{n} = \dbinom{9}{0} + \dbinom{9}{1} + \cdots + \dbinom{9}{9}\] to get 512. However, 0 itself is not a valid case (since it has 1 digit) so we subtract 1. Our answer is $\boxed{\textbf{(E) }511}$.



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 10, since there are more digits to be tested and there are already 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 $\boxed{\textbf{(E) }511}$.

~Stead (a.k.a. Aaron)

Solution 5

We start by calculating the number of upnos. Suppose we are constructing an upno of $n$ digits such that $n \ge 2$. An upno can't start with a "$0$", so there are $9$ digits to choose from. There are $\dbinom{9}{n}$ ways to choose an upno with $n$ digits. This is because for each combination of digits, only one combination can form an upno. Therefore, for $2 \le n \le 9$, the total number of upnos is \[\binom{9}{2} + \binom{9}{3} + \binom{9}{4} + \cdots + \dbinom{9}{9} = 2^9 - \binom{9}{1} - \binom{9}{0} = 2^9 - 10.\]

Similarly, the digits of a downo of $n$ digits can be chosen among 10 digits to choose from, since $0$ can be a digit of the downo as the last digit. Thus, the number of downos is \[\binom{10}{2} + \binom{10}{3} + \binom{10}{4} + \cdots + \dbinom{10}{9} + \dbinom{10}{10} = 2^{10} - \binom{10}{1} - \binom{10}{0} = 2^{10} - 11.\] Thus, \[|U - D| = |(2^9 - 10) - (2^{10} - 11)| = (2^{10} - 11) - (2^9 - 10) = 2^{10} - 2^9 - 1 = \boxed{\textbf{(E) }511}\]


Solution 6 (1-1 Correspondence)

Notice that the number of upno integers are the number of subsets of the set $(1, 2, 3, 4, 5, 6, 7, 8, 9)$, excluding the empty set ($\emptyset$) and the one-digit integers. So, the number of upno integers is $2^9-1-9=502$.

The number of downo numbers are similar, but with $0$. So, the number of downo integers is the number of subsets of the set $(0, 1, 2, 3, 4, 5, 6, 7, 8, 9)$, excluding the empty set and the one-digit integers. So, the number of downo integers is $2^{10}-1-10=1013$.

Therefore, the difference between the number of downo integers and upno is $1013-502=\boxed{\textbf{(E) }511}$.


Solution 7 (Bijection)

Observe that each downno (that doesn't end with a $0$) maps to one upno. It suffices to count the number of downnos that end with $0$.

Given that a downno ends with $0$, we want to choose the remaining digits from the set $\{1, 2, \ldots, 9\}$. We can arrange the elements of any subset so that they are increasing. We may include or exclude any of the $9$ elements, giving $2^{9}$ subsets. Subtracting $1$ to account for the empty set, we have $2^9 - 1 = \boxed{\textbf{(E) }511}$ more downnos than upnos.

-Benedict T (countmath1)

Video Solution 1 by OmegaLearn


Video Solution


~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See also

2023 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 10 Problems and Solutions

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