Difference between revisions of "2010 USAJMO Problems/Problem 1"
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− | Thus, we can only match k with <math>a_k</math> if they have the same f value. Thus, to find P(k), we can do it by f value, permuting the a_k with f value 1, then 2, ... Thus, our answer is: | + | Thus, we can only match k with <math>a_k</math> if they have the same f value. Thus, to find P(k), we can do it by f value, permuting the <math>a_k</math> with f value 1, then 2, ... Thus, our answer is: |
<math>P(n) = \prod_{1 \leq i \leq n, g(i) = 1} \left\lfloor \sqrt{\frac{n}{i}} \right \rfloor !</math> | <math>P(n) = \prod_{1 \leq i \leq n, g(i) = 1} \left\lfloor \sqrt{\frac{n}{i}} \right \rfloor !</math> |
Latest revision as of 20:23, 5 April 2021
Contents
Problem
A permutation of the set of positive integers is a sequence such that each element of appears precisely one time as a term of the sequence. For example, is a permutation of . Let be the number of permutations of for which is a perfect square for all . Find with proof the smallest such that is a multiple of .
Solutions
We claim that the smallest is .
Solution 1
Let be the set of positive perfect squares. We claim that the relation is an equivalence relation on .
- It is reflexive because for all .
- It is symmetric because .
- It is transitive because if and , then , since is closed under multiplication and a non-square times a square is always a non-square.
We are restricted to permutations for which , in other words to permutations that send each element of into its equivalence class. Suppose there are equivalence classes: . Let be the number of elements of , then .
Now . In order that , we must have for the class with the most elements. This means , since no smaller factorial will have as a factor. This condition is sufficient, since will be divisible by for , and even more so .
The smallest element of the equivalence class is square-free, since if it were divisible by the square of a prime, the quotient would be a smaller element of . Also, each prime that divides divides all the other elements of , since and thus . Therefore for all . The primes that are not in occur an even number of times in each .
Thus the equivalence class . With , we get the largest possible . This is just the set of squares in , of which we need at least , so . This condition is necessary and sufficient.
Solution 2
This proof can also be rephrased as follows, in a longer way, but with fewer highly technical words such as "equivalence relation":
It is possible to write all positive integers in the form , where is the largest perfect square dividing , so is not divisible by the square of any prime. Obviously, one working permutation of is simply ; this is acceptable, as is always in this sequence.
Lemma 1. We can permute any numbers that, when each divided by the largest perfect square that divides it, yield equal quantities .
Proof. Let and be the values of and , respectively, for a given as defined above, such that is not divisible by the square of any prime. We can obviously permute two numbers which have the same , since if where and are 2 values of , then , which is a perfect square. This proves that we can permute any numbers with the same value of .
End Lemma
Lemma 2. We will prove the converse of Lemma 1: Let one number have a value of and another, . and are both perfect squares.
Proof. and are both perfect squares, so for to be a perfect square, if is greater than or equal to , must be a perfect square, too. Thus is times a square, but cannot divide any squares besides , so ; . Similarly, if , then for our rules to keep working.
End Lemma
We can permute numbers with the same in ways. We must have at least 67 numbers with a certain so our product will be divisible by 67. Obviously, then it will also be divisible by 2, 3, and 5, and thus 2010, as well. Toms as , in general, we need numbers all the way up to , so obviously, is the smallest such number such that we can get a term; here 67 terms are 1. Thus we need the integers , so , or , is the answer.
Solution Number Sense
We have to somehow calculate the number of permutations for a given . How in the world do we do this? Because we want squares, why not call a number , where is the largest square that allows to be non-square? is the only square can be, which only happens if is a perfect square.
For example, , therefore in this case .
I will call a permutation of the numbers , while the original I will call .
Note that essentially we are looking at "pairing up" elements between and such that the product of and is a perfect square. How do we do this? Using the representation above.
Each square has to have an even exponent of every prime represented in its prime factorization. Therefore, we can just take all exponents of the primes and if there are any odd numbers, those are the ones we have to match- in effect, they are the numbers mentioned at the beginning.
By listing the values, in my search for "dumb" or "obvious" ideas I am pretty confident that only values with identical s can be matched together. With such a solid idea let me prove it.
If we were to "pair up" numbers with different s, take for example with an of and with an of , note that their product gives a supposed of because the values cancel out. But then, what happens to the extra left? It doesn't make a square, contradiction. To finish up this easy proof, note that if a "pair" has different values, and the smaller one is , in order for the product to leave a square, the larger value has to have not just but another square inside it, which is absurd because we stipulated at the beginning that was square-free except for the trivial multiplication identity, 1.
Now, how many ways are there to do this? If there are numbers with , there are clearly ways of sorting them. The same goes for by this logic. Note that the as stated by the problem requires a thrown in there because , so there has to be a with 67 elements with the same . It is evident that the smallest will occur when , because if is bigger we would have to expand to get the same number of values. Finally, realize that the only numbers with are square numbers! So our smallest , and we are done.
I relied on looking for patterns a lot in this problem. When faced with combo/number theory, it is always good to draw a sketch. Never be scared to try a problem on the USAJMO. It takes about 45 minutes. Well, it's 2010 and a number 1. Cheers!
-expiLnCalc
Solution 4
We can factor all positive integers as , where are positive integers and is divisible by no square other than 1.
Break up the set into subsets. Let subset contain numbers of the form , where remains constant.
Now, consider a valid arrangement. Take all the numbers from subset and their indices. We claim these numbers can be permutated over the indices, to create another valid arrangement. But this is trivial; if two numbers and were to swap places, it is clear that multiplying by indices would create perfect squares once more.
Thus, the number of permutations in total is the product of those in each set.
In order for this total to be divisible by , we only need for it to divide . So, at least one subset must have elements. To minimize , let the subset contain elements of the form . Then ranges from , so the largest element is . Setting to this value, the answer is .
~Leonard_my_dude~
Solution 5
It's well known that there exists and such that , no square divides other than 1, and is a perfect square.
Lemma: is a perfect square if and only if
We prove first: If , is a perfect square.
, which is a perfect square.
We will now prove: If is a perfect square, .
We do proof by contrapositive: If , is not a perfect square.
is the p-adic valuation of k. (Basically how many factors of p you can take out of k)
Note that if , By the Fundamental Theorem of Arithmetic, and 's prime factorization are different, and thus there exists a prime p, such that . Also, since and is squarefree, . Thus, , making not a square.
End Lemma
Thus, we can only match k with if they have the same f value. Thus, to find P(k), we can do it by f value, permuting the with f value 1, then 2, ... Thus, our answer is:
For all , doesn't have a factor of 67. However, if , the first term will be a multiple of 2010, and thus the answer is
-Alexlikemath
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
- <url>viewtopic.php?t=347303 Discussion on AoPS/MathLinks</url>
2010 USAJMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.