**From New Scientist #1629, 8th September 1988** [link]

In both the football league table and addition sum below, letters have been substituted for digits. Each letter stands or should stand for a different digit (from 0 to 9), and different letters should stand for different digits. And so they do except for the fact that one of the letters is incorrect on one of the occasions on which it appears (if indeed it appears more than once).

The three teams are eventually going to play each other once, or perhaps they have already done so.

Which letter was wrong? What should it be?

Find the scores in the football matches and write the addition sum out with numbers substituted for letters.

[enigma478]

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There are at most 3 matches played (A v B, A v C, B v C).

If we suppose the “played” and “points” columns are correct.

If no matches had been played, all the values would be 0, so there would be more than one mistake in the table (and in particular the columns we are supposing are correct).

If all matches had been played, then every value in the “played” column would be 2, so there would be a mistake in the “played” column.

If only one match had been played then

kandgwould be0and1(in some order). But they appear in the other order in the “points” column. So one team would have to have played 0 matches and gained 1 point. This is impossible.If two matches had been played. Then one team has played 2 matches, and the other teams have played 1 match each. So

kandgwould be1and2in some order. In each of the 2 matches 2 points are awarded, so the sum of the “points” column is 4, somwould have to be1, but we know one ofgorkis1. So this is not possible either.So there is no scenario where both the “played” and “points” columns are correct, so the error must be in one of these columns.

This Python program uses the [[

Football()]] helper class from theenigma.pyto find the erroneous entry in the table, and the outcomes of the matches. It runs in 83ms.Run:[ @repl.it ]Solution:The points entry for C should bek(= 1), notm(= 0). The scores in the matches are: A v B = 4 – 3; A v C = not played; B v C = 4 – 4. The addition sum is: 3 + 4 = 7.