2025 AMC 12A Problems/Problem 23: Difference between revisions
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==Problem 23== | |||
==Solution 1== | |||
To satisfy the conditions, a <imath>\textit{fair}</imath> integer must have no digit be a local minimum. Let's say we have <imath>n</imath> distinct digits, with each digit being a number from <imath>1</imath> to <imath>9</imath>. To create a <imath>\textit{fair}</imath> integer, we begin by placing the largest digit. For the second-largest digit, we can either place this digit to the right or to the left of the string already created. We have these <imath>2</imath> options for the third-largest digit, and so on. Therefore, there are <imath>2^{n-1}</imath> valid permutations to create a <imath>\textit{fair}</imath> integer. | To satisfy the conditions, a <imath>\textit{fair}</imath> integer must have no digit be a local minimum. Let's say we have <imath>n</imath> distinct digits, with each digit being a number from <imath>1</imath> to <imath>9</imath>. To create a <imath>\textit{fair}</imath> integer, we begin by placing the largest digit. For the second-largest digit, we can either place this digit to the right or to the left of the string already created. We have these <imath>2</imath> options for the third-largest digit, and so on. Therefore, there are <imath>2^{n-1}</imath> valid permutations to create a <imath>\textit{fair}</imath> integer. | ||
Revision as of 13:39, 6 November 2025
Problem 23
Solution 1
To satisfy the conditions, a 
integer must have no digit be a local minimum. Let's say we have 
distinct digits, with each digit being a number from 
to 
. To create a integer, we begin by placing the largest digit. For the second-largest digit, we can either place this digit to the right or to the left of the string already created. We have these 
options for the third-largest digit, and so on. Therefore, there are 
valid permutations to create a integer.
We must also choose which digits will be in the permutation. If you are creating an -digit long integer, there are 
ways to pick which digits will be in the number.
Therefore, for each 
, the number of fair integers of length is:
![\[\binom{9}{n} \cdot 2^{n-1}.\]](http://latex-new.aopstest.com/a/1/2/a12ededc3f03a65e075762f9da27893cec04cbd0.png)
Summing over all :
![\[\sum_{n=1}^9{\binom{9}{n} \cdot 2^{n-1}}\]](http://latex-new.aopstest.com/6/e/5/6e511341d815ff5a611d17f592f5332bef75eeb8.png)
![\[=\frac{1}{2}\left(\sum_{n=0}^9{\binom{9}{n}}2^n -1\right)\]](http://latex-new.aopstest.com/a/c/8/ac85df3ce74ad8b87d6e34a1902e290438202057.png)
![\[=\frac{1}{2}\left((1+2)^9 -1 \right) = \frac{1}{2}(19682) = \boxed{9841}.\]](http://latex-new.aopstest.com/d/9/0/d9039135a5659bfd2cd16be651f0f151c05a5e79.png)
Note that the Binomial Theorem was used to equate
![\[\sum_{n=0}^9{\binom{9}{n}}2^n = (1+2)^9.\]](http://latex-new.aopstest.com/d/9/9/d99da1d5c23d62fe6bbd15ebf43228c105f7d7da.png)
~lprado
