2023 AMC 10B Problems/Problem 16

Revision as of 16:24, 15 November 2023 by Technodoggo (talk | contribs) (Solution (pending))

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