Difference between revisions of "2010 USAJMO Problems/Problem 1"
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We can permute <math>l</math> numbers with the same <math>p</math> in <math>l!</math> ways. We must have at least 67 numbers with a certain <math>p</math> 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 <math>h</math>, in general, we need numbers all the way up to <math>h\cdot 67^2</math>, so obviously, <math>67^2</math> is the smallest such number such that we can get a <math>67!</math> term; here 67 <math>p</math> terms are 1. Thus we need the integers <math>1, 2, \ldots, 67^2</math>, so <math>67^2</math>, or <math>\boxed{4489}</math>, is the answer. | We can permute <math>l</math> numbers with the same <math>p</math> in <math>l!</math> ways. We must have at least 67 numbers with a certain <math>p</math> 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 <math>h</math>, in general, we need numbers all the way up to <math>h\cdot 67^2</math>, so obviously, <math>67^2</math> is the smallest such number such that we can get a <math>67!</math> term; here 67 <math>p</math> terms are 1. Thus we need the integers <math>1, 2, \ldots, 67^2</math>, so <math>67^2</math>, or <math>\boxed{4489}</math>, is the answer. | ||
− | + | ==Solution Number Sense== | |
− | + | We have to somehow calculate the number of permutations for a given <math>n</math>. How in the world do we do this? Because we want squares, why not call a number <math>k=m*s^2</math>, where <math>s</math> is the largest square that allows <math>m</math> to be non-square? For example, <math>126 = 14 * 3^2</math>, therefore in this case <math>k=126, m = 14, s = 3</math>. | |
{{alternate solutions}} | {{alternate solutions}} | ||
Revision as of 12:29, 5 April 2018
Contents
[hide]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? For example, , therefore in this case . 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 | ||
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