2010 USAJMO Problems/Problem 1
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) | ||
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Followed by Problem 2 | |
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