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Prove that from 17 different integers you can always choose 5 so the sum will be divisible by 5.

I tried with positive,negative numbers. Even, odd numbers etc but can't find the solution. Any thoughts on what should the holes be ?

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"Divisible by 5" suggests considering modulo 5. Can you have 5 of them in the same residue class mod 5? If not, will you have at least 1 in each residue class? If so, notice that $0+1+2+3+4 = 10$.

As chubakueno pointed out, this isn't a tight bound. We can easily cut it down by the following.

We can add 1 to each number and it doesn't affect whether there are 5 of them with the property, so we can add until residue class 0 is the biggest.

If there are at least 3 "0"s, there cannot be both "1" and "4" or both "2" and "3", giving at most $4+4+4 = 12$. In the other case there are at most 2 "0"s, and hence at most 2 in each residue class, giving at most 10 numbers in total. This gives an upper bound of 12, which still isn't tight, and if we want we can push even further.

If there are at least 3 "0"s,

  There cannot be:

    "1,2,2" or "1,1,1,2"

    "1,1,3" or "1,3,3,3"

    "2,4,4" or "2,2,2,4"

    "3,3,4" or "3,4,4,4"

  Therefore in any pair of non-empty non-zero residue classes has at most 3 numbers in total

  But as before for every non-zero residue class its complementary class must be empty

  So there will be at most 2 non-empty non-zero residue classes

  If there are 2 non-empty non-zero residue classes,

    They have at most 3 numbers in total

  If there is at most 1 non-empty residue class,

    There are at most 4 numbers in it

  Therefore there are at most $4+4 = 8$ numbers

If there are at most 2 "0"s,

  If there is a non-zero residue class with 2 numbers in it,

    We could make it such that it is either residue 1 or residue 2

    If we have 2 "0"s and 2 "1"s,

      There cannot be any "3" and there can be at most 1 "2" and 1 "4"

      Therefore there are at most $2+2+1+1 = 6$ numbers

    If we have 2 "0"s and 2 "2"s,

      There cannot be any "1" and there can be at most 1 "3" and 1 "4"

      Therefore there are at most $2+2+1+1 = 6$ numbers

  If every non-zero residue class has at most 1 number in it,

    There are at most $2+1+1+1+1 = 6$ numbers

  Therefore in any case there are at most 6 numbers

Therefore there are at most 8 numbers and it is tight because of $(0,0,0,0,1,1,1,1)$

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  • $\begingroup$ I was mistaken. Had to edit. Thank you ! $\endgroup$ – lvi Feb 20 '14 at 16:52
  • $\begingroup$ @lvi his hint uses that $16<17$. But actually, 17 is not a tight bound. $\endgroup$ – chubakueno Feb 20 '14 at 16:54
  • $\begingroup$ @chubakueno: Please help me check my proof for the tight bound, in case I made some mistake! $\endgroup$ – user21820 Feb 21 '14 at 6:00
  • $\begingroup$ I didn't understand Therefore in any pair of non-empty non-zero residue classes has at most 3 numbers in total .Is there a typo, maybe? $\endgroup$ – chubakueno Feb 21 '14 at 23:04
  • $\begingroup$ It's because those blocking combinations between pairs of classes are such that if you have 1 in one class, you can have at most 2 in the other class. For example if you have 1 "1", you can have at most 1 "2", and if you have 1 "2", you can have at most 2 "1"s. $\endgroup$ – user21820 Feb 22 '14 at 2:35
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Since you don't seem to be familiar with modular arithmetic I won't use the language of congruences.

Every integer is of one of the following forms, $$ 5k, \ 5k + 1, 5k + 2, 5k + 3, 5k + 4; \; \text{for some integer $k$} $$

Now, if you pick $17$ integers by the pigeonhole principle either you must pick at least $1$ from each class or else you will have $5$ from the same class.

Say you picked at least $1$ from each class then they are of the form, $$5k_1, \ 5k_2 + 1, \ 5k_3 + 2, \ 5k_4 + 3, \ 5k_5 + 4$$ Now their sum is equal to $$5(k_1 + k_2 + k_3 + k_4 + k_5) + 10 \;\; \text{which is clearly divisible by $5$}$$

OR

You've picked $5$ integers from the same class and they are of the form $$5k_1 + i, \ 5k_2 + i, \ 5k_3 + i, \ 5k_4 + i, \ 5k_5 + i$$ where $i \in \{0,1,2,3,4\}$. It is easily seen that their sum too is divisible by $5$

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    $\begingroup$ I think that should be "... if you pick $17$ integers by the pigeonhole principle you must pick $5$ from the same class or at least 1 from each class..." $\endgroup$ – Christopher Feb 20 '14 at 16:18
  • $\begingroup$ @user73985: You're right.. Corrected.. $\endgroup$ – Ishfaaq Feb 20 '14 at 16:25
  • $\begingroup$ @Ishfaaq understand if the 5 selected integers are in either the same class or if they are selected so that at least one is in each class we can find their sum to be divisible by zero.But i don't understand how we get to think as` if you pick 17 integers by the pigeonhole principle either you must pick at least 1 from each class or else you will have 5 from the same class.` $\endgroup$ – clarkson Nov 16 '14 at 8:24

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