The pigeonhole hole principle

[Albi]

Show that in every subset with 7 elements, from the set of Integers, there exist at least two different numbers x and y, whose sum or difference is divisible by 10.

I know that this has to do with the pigeonhole principle, but I don't know how to use it in this problem, can someone help me?

 

[lev888]

Show that in every subset with 7 elements, from the set of Integers, there exist at least two different numbers x and y, whose sum or difference is divisible by 10.

I know that this has to do with the pigeonhole principle, but I don't know how to use it in this problem, can someone help me?

Can you describe the principle and post an example of a proof that uses it?

 

[Albi]

Can you describe the principle and post an example of a proof that uses it?

The pigeonhole principle: If n objects are placed in k boxes, where n > k , then there is at least one box containing at least [imath]\left \lceil \frac{n}{k} \right \rceil[/imath] objects.

 

[pka]

Show that in every subset with 7 elements, from the set of Integers, there exist at least two different numbers x and y, whose sum or difference is divisible by 10.
I know that this has to do with the pigeonhole principle, but I don't know how to use it in this problem, can someone help me?

If [imath]\mathscr{A}[/imath] is a subset of seven integers then there are [imath]\binom{7}{2}=21[/imath] two element subsets of [imath]\mathscr{A}[/imath] At this point, were I you, I would make-up a subset of seven integers. Take those seven two at time a time, there are only twenty-one pairs.
Add each pair together and then subtract the two. If your problem is stated correctly the some result is a multiple of ten.
[imath](x+y)\equiv 0\mod(10)\text{ or }(x-y)\equiv 0\mod(10)[/imath] for some pair.

 

[Albi]

If [imath]\mathscr{A}[/imath] is a subset of seven integers then there are [imath]\binom{7}{2}=21[/imath] two element subsets of [imath]\mathscr{A}[/imath] At this point, were I you, I would make-up a subset of seven integers. Take those seven two at time a time, there are only twenty-one pairs.
Add each pair together and then subtract the two. If your problem is stated correctly the some result is a multiple of ten.
[imath](x+y)\equiv 0\mod(10)\text{ or }(x-y)\equiv 0\mod(10)[/imath] for some pair.

I formed the pairs you said and I got 3 pairs that are congruent modulo 10, so the condition is satisfied, but how can I generalise the solution for every subset of Integers with 7 elements?

 

[pka]

I formed the pairs you said and I got 3 pairs that are congruent modulo 10, so the condition is satisfied, but how can I generalise the solution for every subset of Integers with 7 elements?

Now is is your turn to do some mathematics. if the sum of two integers is a multiple of ten what can be said of the two?
If the difference of two integers is a multiple of ten what can be said of the two?
How does the answers of the above relate to the number seven?
Remember, you are proving, in any collection of seven integers you have at least two the sum or difference of which is a multiple of ten.

 

[Albi]

Now is is your turn to do some mathematics. if the sum of two integers is a multiple of ten what can be said of the two?
If the difference of two integers is a multiple of ten what can be said of the two?
How does the answers of the above relate to the number seven?
Remember, you are proving, in any collection of seven integers you have at least two the sum or difference of which is a multiple of ten.

When I divide a number by 10 there are 10 possible remainders:{0,1,2,3,4,5,6,7,8,9}, I notice that 1+9 = 2+8 = 3+7 = 4+6= 10, meaning the numbers that give a remainder of 1 and 9, when divided by 10, are divisible by 10, etc, and there are the multiples of 10
If two integers give the same remainder when divided by ten, then their difference will be divisible by 10 .
I don't know if I can consider the pairs {1,9}, {2,8), {3,7}, {4,6}, and the muttiples of ten, as the the pigeonholes, and the seven integers from the subset as the pigeons, and then use [imath]\left \lceil \frac{7}{5} \right \rceil=2[/imath].