2022 AMC 10A Problems/Problem 24
- The following problem is from both the 2022 AMC 10A #24 and 2022 AMC 12A #24, so both problems redirect to this page.
Contents
- 1 Problem
- 2 Solution 1 (Parking Functions)
- 3 Solution 2 (Casework)
- 4 Solution 3 (Recursive Equations Approach)
- 5 Solution 4 (Fake solve, incorrect logic, correct answer by coincidence)
- 6 Solution 5 (Mod 5 Trick Redeemed)
- 7 Solution 6 (Casework Bash)
- 8 Video Solution by Math-X (Smart and Simple)
- 9 Video Solution
- 10 Video Solution
- 11 Video Solution By OmegaLearn using Complementary Counting
- 12 See Also
Problem
How many strings of length formed from the digits , , , , are there such that for each , at least of the digits are less than ? (For example, satisfies this condition because it contains at least digit less than , at least digits less than , at least digits less than , and at least digits less than . The string does not satisfy the condition because it does not contain at least digits less than .)
Solution 1 (Parking Functions)
For some , let there be parking spaces counterclockwise in a circle. Consider a string of integers each between and , and let cars come into this circle so that the th car tries to park at spot , but if it is already taken then it instead keeps going counterclockwise and takes the next available spot. After this process, exactly one spot will remain empty.
Then the strings of numbers between and that contain at least integers for are exactly the set of strings that leave spot empty. Also note for any string , we can add to each (mod ) to shift the empty spot counterclockwise, meaning for each string there exists exactly one with so that leaves spot empty. This gives there are such strings.
Plugging in gives such strings.
~oh54321
Note: If you are confused, the problem is equivalent to “How many strings of length formed from the digits , , , , , are there such that for each , at least of the digits are less than ?”
~inaccessibles
Solution 2 (Casework)
Note that a valid string must have at least one
We perform casework on the number of different digits such strings can have. For each string, we list the digits in ascending order, then consider permutations:
- The string has different digit.
- The string has different digits.
- The string has different digits.
- The string has different digits.
- The string has different digits.
The only possibility is
There is string in this case.
We have the following table: There are strings in this case.
We have the following table: There are strings in this case.
We have the following table: There are strings in this case.
There are strings in this case.
Together, the answer is
~MRENTHUSIASM
Solution 3 (Recursive Equations Approach)
Denote by the number of -digit strings formed by using numbers , where for each , at least of the digits are less than .
We have the following recursive equation: and the boundary condition for any .
By solving this recursive equation, for and , we get
For and , we get
For and , we get
For and , we get
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 4 (Fake solve, incorrect logic, correct answer by coincidence)
The number of strings is as shown by Solution 1 (Parking Function), which is always equivalent to 1 (mod n). Thus you can choose .
However, you **CANNOT** prove that fact from this *incorrect* argument:
Let the set of all valid sequences be . Notice that for any sequence in , the 5 cyclic permutations must also belong in . However, one must consider the edge case all 5 elements are the same (only ), in which case all sequences listed are equivalent.
Alas, one must also consider cases like 10101 that only have 3 distinct cyclic permutations, so we cannot get a useful constraint from this view, unless *all* the cases are categorized and counted.
Thus it is NOT implied that , so it is not justified to choose by inspection of answer options.
If you solve for smaller by bashing, you can find , , , , which leads to an Engineer's Induction guess that , or the exact formula
~ idea by Tau, logic corrected by oinava
Solution 5 (Mod 5 Trick Redeemed)
Solution 4 tried to observe the answer modulo to easily solve the problem, but apparently had faulty logic. This solution is still completely viable though:
Notice that for any valid set , if there is at least one element in the set that is unique (i.e. there is at least one digit in the set that is found nowhere else in the set), then the number of distinct permutations of the set is clearly divisible by (alternatively there is no factor of 5 in the denominator of the multinomial coefficient unless an element is repeated 5 times). Therefore to evaluate the answer modulo , we only need to look at sets where each element has a multiplicity of at least (i.e. appears twice or more in the set).
These sets are of the form and . The first set can be permuted in , and the second set can be permuted one way, and the only set of the form is . Therefore the answer is congruent to and you CAN choose .
~SpencerD.
Solution 6 (Casework Bash)
We list out all possibilities for the string, sorted in increasing order. Note that 1 cannot appear before the 2nd position, 2 cannot appear before the 3rd position, and so on. We also make a list of the number of possible permutations of each sorted string. 00000 - 1
00001 - 5
00002 - 5
00003 - 5
00004 - 5
00011 - 10
00012 - 20
00013 - 20
00014 - 20
00022 - 10
00023 - 20
00024 - 20
00033 - 10
00034 - 20
00111 - 10
00112 - 30
00113 - 30
00114 - 30
00122 - 30
00123 - 60
00124 - 60
00133 - 30
00134 - 60
00222 - 10
00223 - 30
00224 - 30
00233 - 30
00234 - 60
01111 - 5
01112 - 20
01113 - 20
01114 - 20
01122 - 30
01123 - 60
01124 - 60
01133 - 30
01134 - 60
01222 - 20
01223 - 60
01224 - 60
01233 - 60
01234 - 120
Finally, adding all of these up, we obtain . (Or alternatively, we could notice as in solution 5 that the only possible string with only 1 permutation is 00000, and thus the answer must be congruent to 1 modulo 5)
The key to organizing all of these cases is to increase the digits one by one, starting from the right and iterating through all the possible cases.
~ MathHayden
Video Solution by Math-X (Smart and Simple)
https://youtu.be/7yAh4MtJ8a8?si=aBXOzKzoAZwkb4l1&t=9314
~Math-X
Video Solution
~MathProblemSolvingSkills.com
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution By OmegaLearn using Complementary Counting
~ pi_is_3.14
See Also
2022 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 |
2022 AMC 12A (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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.