Subsets and divisibility

Consider a set of distinct positive integers. What is the minimum number of integers required in the set to guarantee that, regardless of what the integers are, there is at least one subset of nine integers whose sum is divisible by 9?


The answer is 17.

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1 solution

Denton Young
Mar 1, 2017

First, we show that 17 integers works.

Lemma: Given any five integers, there is at least one subset of three integers whose sum is divisible by 3.

Proof: Let us try to find a set of five integers that contains no subset of three integers whose sum is divisible by 3.. Each of the five integers is, for our purposes, equal to 0, 1, or 2, because we are concerned only with divisions by 3. We cannot have one of each of these, because 0 + 1 + 2 is divisible by 3. We must therefore have at most two of the types. But the pigeonhole principle then implies that we have at least three of one of the types. The sum of these three integers is divisible by 3.

Returning to the original problem, pick five integers to obtain a triplet whose sum is divisible by 3. Then pick another five integers to obtain another such triplet. We can continue to do this for a total of five times, given the seventeen integers.

We now have five triplets, each of whose sum is divisible by 3. As far as divisions by 9 are concerned, these sums are equal to 0, 3, or 6. We can now use the same reasoning as in the lemma above (but with everything scaled up by a factor of 3) to show that we can find a set of three triplets that has a sum divisible by 9. In other words, we have found a set of nine integers whose sum is divisible by 9.

Next, we show that 16 integers is insufficient.

Consider the set {9,10, 18,19, 27, 28, 36, 37, 45, 46, 54, 55, 63, 64, 72, 73}, containing 16 integers. 8 of them are divisible by 9 and 8 have a remainder of 1 when divided by 9. If you select 9 integers from this set, you will end up with at least 1 and at most 8 of the integers which have a remainder of 1 divisible by 9, so will not end up with a set with a sum divisible by 9.

What if I choose a subset which has all the multiples of nine?

Vidit Kulshreshtha - 4 years, 3 months ago

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Then you would have 8 elements that were multiples of 9, plus one element that wasn't, and your sum would not be a multiple of 9.

Denton Young - 4 years, 3 months ago

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@Denton Young you have not got my doubt right. I'm saying that if I have a set of 9 elements that have all the multiples of nine. And every set is a subset of itself. So my work would be done with 9 elements only.

Vidit Kulshreshtha - 4 years, 3 months ago

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@Vidit Kulshreshtha I should rephrase the problem then. What is the least number you need to GUARANTEE for any set of [that number] of integers, regardless of what the integers are?

Denton Young - 4 years, 3 months ago

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@Denton Young That is something at which I can agree..

Vidit Kulshreshtha - 4 years, 3 months ago

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@Vidit Kulshreshtha Thank you for your help. The problem is now correctly phrased.

Denton Young - 4 years, 3 months ago

So only 9 elements will do

Vidit Kulshreshtha - 4 years, 3 months ago

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Correct. Your subset has to have nine elements in it, no more and no less.

Denton Young - 4 years, 3 months ago

i did the same thing. 25 integers in ascending order. first five numbers have three numbers such that sum is divisible by 3. let sum=3n1. for next 5 we have 3n2, considering n1,n2,n3,n4,n5 we have three numbers such that sum of them is divisible by 3. so 25 integers works.

Srikanth Tupurani - 2 years, 3 months ago

i am sorry i made a mistake. if we have have six distinct numbers. a1<a2<a3<a4<a5<a6 then we need to consider a1.a2,a3,a4,a5 and a2,a3,a4,a5,a6.

Srikanth Tupurani - 2 years, 3 months ago

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