2023 AMC 10B Problems/Problem 16

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.

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2023 AMC 10B Problems/Problem 16

Solution

Solution