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?
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.
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.