# Pigeon hole principle with sum of 5 integers

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 $(0,0,0,0,1,1,1,1)$

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 $5$