2025 AMC 10A Problems/Problem 24: Difference between revisions
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== Solution 4 (Cheese 🧀)== | == Solution 4 (Cheese 🧀)== | ||
Notice that answer choices A and B are way too small, and answer choices D and E are way too big. This leaves us with <cmath>\boxed{(C) 9841}</cmath>. | Notice that answer choices A and B are way too small, and answer choices D and E are way too big. This leaves us with <cmath>\boxed{\text{(C) } 9841}</cmath>. | ||
Note: Do not try this on an actual test unless you have to. This is a really risky solution. | Note: Do not try this on an actual test unless you have to. This is a really risky solution. | ||
Revision as of 19:51, 7 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
Solution 2
There is one very interesting detail about fair positive integers: The second a digit becomes less than its preceding one, the rest of the digits following must also be in decreasing order. This implies that every fair positive integer increases, reaches a maximum digit, and then decreases. This type of permutation of numbers is called a unimodal permutation.
Let's say we have a set of 
distinct integers. We want to find the number of unimodal permutations of the set. First, we choose the integers to be in the increasing part of the permutation. For each of the integers, they are either in the increasing sequence or not. This gives 
possible increasing sequences, and the decreasing sequence is decided from the remaining integers. But, we must divide this count by 
as we do not need to count the choices for our vertex (the vertex is in both the increasing and decreasing sequence). Hence, the number of unimodal permutations of a set of length is 
Therefore, for the number of 
digit fair positive integers, there are 
ways to choose the digits, and 
ways to make a unimodal permutation with those digits. Our total count is then
![\[\sum_{i=1}^{9}\dbinom{9}{i}2^{i-1}=\dfrac{1}{2}\sum_{i=1}^{9}\dbinom{9}{i}2^{i}=\dfrac{1}{2}((1+2)^9-1)=\boxed{\text{(C) }9841}.\]](http://latex-new.aopstest.com/b/c/1/bc16e4c126a068e53a3952889851456eb45be5a2.png)
~Tacos_are_yummy_1
Alternate Counting Method
Or, to calculate ![\[\sum_{i=1}^{9}\dbinom{9}{i}2^{i-1},\]](http://latex-new.aopstest.com/8/1/9/819cea25f7c5c697a35826b89f24aa0e22cfb1a3.png)
we can double it and then half it to create a more symmetric structure, giving us ![\[\frac{1}{2} \sum_{i=1}^{9}\dbinom{9}{i}2^{i}.\]](http://latex-new.aopstest.com/0/5/c/05c1f27daf96c42a00e68f136c3bdd07087d14b8.png)
Consider a counting argument in which there are 
people, and you choose a nonempty group of them to respond, and they can say either vote yes or no to a prompt. Equivalently, this would be identical to giving each person three choices, to not vote, to vote yes, or to vote no. Therefore, the aforementioned sum is equivalent to ![\[\sum_{i=1}^{9}\dbinom{9}{i}2^{i} = 3^i - 1\]](http://latex-new.aopstest.com/7/5/3/7532a10e7ee783a3de8fb5630ff19c3c7d69a6e2.png)
where the -1 comes from when no one votes. Halving that yields 
Solution 3 (Easily Motivated)
Similar to Solution 1, observe that the digits of all "fair" numbers go up and go down. We simply do casework on the peak digit; if it is 
, for example, each of the digits 
can either be to the left of , to the right of , or don't appear in the number at all. This would yield 
such numbers. Sweeping through all possible values of the peak digit from 1 through 9, we get ![\[3^0+3^1+...+3^8=\dfrac{3^9-1}{2}=\boxed{\text{(C) }9841}\]](http://latex-new.aopstest.com/8/4/d/84d66ba81684b61ad5ea7e35e7dabe264c08456e.png)
~Mathkiddie
Solution 4 (Cheese 🧀)
Notice that answer choices A and B are way too small, and answer choices D and E are way too big. This leaves us with ![\[\boxed{\text{(C) } 9841}\]](http://latex-new.aopstest.com/9/8/0/980f6ab36a82a01bc83858e18fc5776f2de09c39.png)
.
Note: Do not try this on an actual test unless you have to. This is a really risky solution.
~Aeioujyot
Video Solution (In 2 Mins)
https://youtu.be/hYZkrBtIFvI?si=dMOaGsXSpxYV17PO ~ Pi Academy
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=dAeyV60Hu5c
See Also
| 2025 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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