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The scores were 6-2-1-0-0-0-1-0-0-0. (There are six 0s, two 1s, one 2, no 3s, no 4s no 5s, one 6, no 7s, no 8s, no 9s.) This is the only solution, as Edward DeLorenzo proved in 1968. A little thought shows that at most one of the digits 5, 6, 7, 8, 9 can occur, and it can occur only once, otherwise there would be too many digits. So the leftmost digit is 5 or more, and there have to be at least five 0s. The analysis then breaks down into several distinct cases, depending on the leftmost digit. The answer given is the only possibility that survives.