2025 AMC 10A Problems/Problem 24: Difference between revisions
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==Solution 1== | ==Solution 1== | ||
Note that the condition that no digit is adjacent to two greater digits means that the digits must be in ascending order up to some largest digit, when they then switch to descending order. To find the number of such numbers, consider two "parts" of the number: the ascending part and the descending part. Note that since each of these has a unique ordering, we just need to divide the digits <imath>1-9</imath> among these parts. For each such digit, it can either be in the ascending part, descending part, or not be in the number, giving <imath>3</imath> possibilities per digit for a total of <imath>3^9</imath> possibilities. However, we can see that the number cannot be "empty", so we have to subtract 1. Furthermore, the largest digit can be on either part, as it will not make a difference. For example, <imath>136\bar742=1367\bar42.</imath> Therefore, we divide by <imath>2</imath> to get <imath>(3^9-1)/2=19682/2=\boxed{\textbf{(C) } 9841}</imath> | Note that the condition that no digit is adjacent to two greater digits means that the digits must be in ascending order up to some largest digit, when they then switch to descending order. To find the number of such numbers, consider two "parts" of the number: the ascending part and the descending part. Note that since each of these has a unique ordering, we just need to divide the digits <imath>1-9</imath> among these parts. For each such digit, it can either be in the ascending part, descending part, or not be in the number, giving <imath>3</imath> possibilities per digit for a total of <imath>3^9</imath> possibilities. However, we can see that the number cannot be "empty", so we have to subtract 1. Furthermore, the largest digit can be on either part, as it will not make a difference. For example, <imath>136\bar742=1367\bar42.</imath> Therefore, we divide by <imath>2</imath> to get <imath>(3^9-1)/2=19682/2=\boxed{\textbf{(C) } 9841}</imath> | ||
~krithikrokcs | |||
Revision as of 13:38, 6 November 2025
Call a positive integer fair if no digit is used more than once, it has no 
s, and no digit is adjacent to two greater digits. For example, 
and 
are fair, but 
and 
are not. How many fair positive integers are there?
Solution 1
Note that the condition that no digit is adjacent to two greater digits means that the digits must be in ascending order up to some largest digit, when they then switch to descending order. To find the number of such numbers, consider two "parts" of the number: the ascending part and the descending part. Note that since each of these has a unique ordering, we just need to divide the digits 
among these parts. For each such digit, it can either be in the ascending part, descending part, or not be in the number, giving 
possibilities per digit for a total of 
possibilities. However, we can see that the number cannot be "empty", so we have to subtract 1. Furthermore, the largest digit can be on either part, as it will not make a difference. For example, 
Therefore, we divide by 
to get 
~krithikrokcs
