Decathlon Logic: 100m Walkthrough

Step 1 – Solis had the fastest time, so scored 16 points. (Solis – 16)

Step 2 – The left and right-hand sides were different heats. The slowest person in the right-hand one was ninth – in other words, only one person in the left-hand heat was faster than anyone in the right-hand one. As ninth place is 8 points, it must be in a corner as it is a multiple of four, and on the right-hand side, so Woodrow scored 8. The remaining multiple of four is 12, which must belong to Green. (Woodrow – 08, Green – 12)

Step 3 – Penn must have doubled Howell’s score as Penn was in the heat that did better. As the scores in the right-hand heat were 8 or better, the even numbered scores Penn could possibly have are 8, 10, 12, 14, 16. However, 8, 12 and 16 have already been taken, leaving just 10 and 14. However, 14 equates to third place, which we know is taken by Popoola or Turner, so it Penn must have ten points, and Howell half that – five points. (Howell – 05, Penn – 10).

Step 4 – There are five scores left to place in the right-hand side: 9, 11, 13, 14, 15. We know that no two orthogonally (horizontally or vertically) adjacent numbers can be one apart. As we know the two squares which 15 and 14 occupy (even if we don’t yet know which is which), we are left with three scores – 9, 11 and 13, one of which must go in the spot immediately above 10 (Moon’s score). This can only be 13. So Moon came fourth. (Moon – 13)

Step 5 – There are now four scores left to place on the right-hand side: 9, 11, 14, 15. We know 14 and 15 are in different columns. If we look at the numbers already filled in, the sum for column three is 23, and the sum for column four is 24. If we pair 15 and 11 and 14 and 9 we get 26 and 23 respectively, which won’t make the columns add up to the same total. So the pairs must be 15 and 9, and 14 and 11. To make the columns both add up to the same total (which is 48), 14 and 11 have to go in the third column, and 15 and 9 in the fourth (rightmost column). As we already know from Woodrow’s clue where 14 and 15 will go, we can place the others. So all four numbers can be filled in at once, which is quite satisfying! (Metcalfe – 11, Popoola – 14, Stanley – 09, Turner – 15).

Step 6 – The path through 5, 4, 3, 2 and 1 has to have Cox scoring 3 otherwise the chain would be broken. Also, in two steps it is only possible to reach Lloyd’s cell, so Lloyd must have scored one and come last. Thus Kemp or Glover has to have scored 2, i.e. finished second-to-last, to complete the chain. However, we know Glover beat McDonagh. It is impossible for Glover to finish second-to-last and beat two other athletes (McDonagh and Lloyd), so Glover cannot have scored 2, so Kemp scored 2. (Cox – 03, Kemp – 02, Lloyd – 01)

Step 7 – There are two remaining scores, 7 and 6. Glover beat McDonagh, so Glover scored the seven. (Glover – 07, McDonagh – 06)