Find the largest value of n such that the complementary set of any subset with n elements of { 1 , 2 , . . . , 1 9 8 4 } contains at least two elements that are relatively prime.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Unexpectedly simple solution. I was complicating things.
Let m be the number of elements of the complementary set which means m+n=1984. If we want to find the max value of n then m must be minimum. So we just need to find the minimum number of elements from m where it contains at least 2 elements that are relatively prime. If I take the even number : 2,4,6,....,1984 and one random odd number there must be at least 2 elements that are relatively prime so m min is 993. N max= 1984- m min , N max = 1984-993, N max=991
I am sorry for my grammatical mistake because I am not good at English
I have a really simple solution to this problem.
Well, because we are dealing with complements, let’s consider the complement (nearly) of this problem.
Let’s try to find the smallest n such that it contains atleast two elements that are relatively prime.
Now, the smallest n is actually 993.
Here’s why:
Consider the subset S = { 2 , 4 , ⋯ , 1 9 8 4 } . Clearly, no two elements are co-prime.
Now, if we consider any other subset S, to maintain this property of no two elements being co-prime to each other, there has to exist atleast one pair ( a i , a j ) , i = j where a i , a j ∈ S such that g c d ( a i , a j ) ∣ g c d ( a k , a l ) ∀ 0 ≤ k ≤ l ≤ ∣ S ∣ . Now, quite obviously, ∣ S ∣ will be maximum when g c d ( a i , a j ) = 2 (which is the above set) ⟹ M I N ( n ) = 9 9 3 .
Problem Loading...
Note Loading...
Set Loading...
For any n > 9 9 1 , consider the subset { 1 , 3 , 5 , ⋯ , 1 9 8 3 , 2 , 4 , ⋯ , 2 ( n − 9 9 2 ) } consisting of n elements. The complement of this set consists of only even integers, which are not relatively prime with each other. Thus, we must have n ≤ 9 9 1 .
Now, we show that n = 9 9 1 is, in fact, the maximum possible value of n . Split the set { 1 , 2 , 3 , ⋯ , 1 9 8 4 } into 992 pairs of consecutive integers:
( 1 , 2 ) , ( 3 , 4 ) , ( 5 , 6 ) , ⋯ , ( 1 9 8 3 , 1 9 8 4 ) .
For any subset of size 991, at most 991 of these pairs will have at least one integer in such a subset. This means that there exists a pair of consecutive integers that do not belong in the subset but that does belong in the complement. Since pairs of consecutive integers are always relatively prime, there always exists two integers in the complement that are relatively prime. Therefore, n = 9 9 1 is the maximum possible value of n .