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Unformatted text preview: Solution Week 80 (3/22/04) Nine divisible by 9 First, consider the following simpler problem: Problem: Given any five integers, show that there is at least one subset of three integers whose sum is divisible by 3. Solution: Let us try to find a set of five integers that contains no subset of three integers whose sum is divisible by 3, a task that we will show is impossible. 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 our auxiliary problem 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. This result is a special case of the following theorem. Theorem: Given any 2 n- 1 integers, there is at least one subset of n integers whose sum is divisible by n ....
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