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2023 AMC 10B Problem 16

2023 AMC 10B Problem 16AMC 10 Step by Step

Name: Date: September 13, 2026

Basic Counting · difficulty 3 of 5 · about 4 min · position P16-20

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 $|U-D|$ ?

A)
$512$
B)
$10$
C)
$0$
D)
$9$
E)
$511$

Answer

Key insight

An upno is a subset of {1,...,9} of size at least 2 and a downno a subset of {0,...,9}; the counts are 2^9 - 10 and 2^10 - 11.

Solution

A strictly increasing (or decreasing) number is determined by its set of digits: once the digits are chosen, there is exactly one way to arrange them in increasing order and one way in decreasing order. So counting upnos and downnos is counting sets of digits.

Upnos: the digit $0$ cannot appear, because in increasing order it would be the leading digit. So an upno is a subset of $\{1, 2, \dots, 9\}$ with at least $2$ elements:
$U = 2^9 - 1 - 9 = 502.$

Downnos: $0$ may appear, since in decreasing order it lands at the end (like $8620$ ). A downno is a subset of $\{0, 1, \dots, 9\}$ with at least $2$ elements:
$D = 2^{10} - 1 - 10 = 1013.$

Therefore $|U - D| = 1013 - 502 = 511$ .

The answer is $\boxed{\textbf{(E)}\ 511}$ .

Why this works

When an ordering is forced, objects correspond bijectively to unordered sets, and sets are counted by powers of $2$ . The whole difference between $U$ and $D$ comes from the single digit $0$ , which is legal only at the end of a number; the extra digit doubles the subset count, and the correction for tiny subsets does the rest.

Alternative approach

Every upno on digits $\{1,\dots,9\}$ reverses into a downno without $0$ , so those cancel in $U - D$ . The surplus downnos are exactly the ones ending in $0$ : a nonempty subset of $\{1, \dots, 9\}$ followed by $0$ , which gives $2^9 - 1 = 511$ .

The trap

Forgetting that 0 can end a downno but can never start an upno, which wrongly makes U = D and gives 0.

Common mistakes

Forgetting that 0 can end a downno but can never start an upno, which wrongly makes U = D and gives 0.
Not removing the empty set and the one-digit numbers, so that $2^{10} - 2^9 = 512$ is reported.

Techniques

Map the objects to something easier to count · Count the complement and subtract from the total