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Congratulations
on finding the Solutions page
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LOGICAL LIGHT-UP.
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ALIEN MUTATIONS. A - head square; B - add legs; C - body square; D - remove arms; E - add antennae; F - biped; G - quadruped; H - biped; I - remove antennae; J - add arms; K - head circle; L - head square. | SILLY COW. Beef grab cash. If we "slice" or rearrange "fresh cabbage" into "all fours", that is, three four-letter words we get the solution meaning "cow stole money". |
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SEQUENCE IN THE SAND. Note that "rowed Tainted Angel" can be read as "rotated angle". Also "Short John Sliver" is a play on "Long John Silver". Together with the word "Sequence" in the title, they all support the notion of a sequence of clockwise rotations of 90 degrees from 0-6 times to give ODD LIMP as the solution. |
WORD PLAY. The play was Henry IV (part one) which included "Cage acting as Hotspur". The latter solution appears from the anagram "a post-Chaucer staging" (4,6,2,7). The characters referred to in the ranking system are alphabetic ones. The seven letters of the alphabet are related to the position of Cage's character (letter) as follows : Hanks' 3 below, Olivier 4 below, Stone's 9 below, Foster's 13 below, Everett's 17 below and Kidman's 20 below. Checking the six possible combinations of letter groups in the alphabet only produces a solution with EHINRVY (5,2). | |
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ARMLESS ALIENS. There were 14 Creakies and 56 Rusties. A robot with n arms requires n females and vice-versa, and the number of members is 2n with n squared links. Let the Creaky tribe have x arms each and the Rusty tribe have y. Then the maximum number of links that the Rusty tribe can make after the mutilation is y(2y-1)/2. The number of extra arms each Creaky needs is 24y/2x allowing x(x + 24y/2x) links. Forming the links equation we have 2(y squared) - 49y - 4(x squared) = 0. We must have y=4x so that y - 4x is a factor. A long division shows that the condition for zero remainder is x=7 so y=28. The numbers of members are twice these. |
THE SLUG AND THE SNAIL. The snail reached the finish in 5 seconds. Since they reach the same position D at the same time t we have 2V(t - T) = X + Vt = D. Rearrangement gives V(t - 2T) = X. Any solution where t = 2T + 1 cannot work since then V = X (must be different numbers). For the set of solutions where T=1, t=4,5,6 and D=6V,8V,10V, respectively, only V=1 allows D less than 10 but then V=T=1 (invalid). For T=2, t=6,7 and D=8V,10V, respectively. Again V=1 which gives X=2 for t=6 (T=X invalid) and X=3 for t=7 with D=10, a valid solution. For T=3, t=8 and D=10V. Here, V=1 gives X=2 with D=10, another valid solution. So there are two valid solutions, both with V=1 and D=10, with T=2, t=7 and T=3, t=8. In both cases t - T = 5. |
| FISH FEAST. The route was XECDBAX as shown which becomes the fish at X enlarged after eating the others! The other possible route XBCDEAX, showing the fish swimming to the right is invalid since the direction is not with the shoal. |  |
CONNECT FOUR. The odd one is detain, the rest can be expressed in letters. | |||
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| THE GREATEST HUMAN BEING. Using (1), Albert is at 1 or 2. If he is at 1, then from (3), Charles is at 2 or 3. However, with Charles at 2 (and Darling at 3), from (2), Curious can only be at 4 and so it is not possible to find just one correct entry in the Surname column (Newtune and Eyeline must be both correct or both incorrect). If Charles is at 3 (with Darling at 4), then from (2), Curious can only be at 3. This gives both Curious and Darling correct (invalid). So Albert must be at 2 (with the bricklayer at 4). From (3), Charles can be at 1 or 3. However, if Charles is at 3 (with Darling at 4), then both Curious and Darling are correct (invalid). So Charles is at 1 (with Darling at 2). From (2), Curious can now be at 3 or 4, but if Curious is at 3 (with the welder at 1) then both the welder and bricklayer are correct (invalid). |
This gives Curious at 4 (with the welder at 2). For
the 1st Name, only Albert is correct at 2, Marie can only be at 4 and so
Isaac is at 3. In the Surname column, only Newtune can be correct at 1,
so Eyeline is at 3. The bricklayer must be the correct speciality at 4,
the cleaner can only be at 1 and so the gardener is at 3.
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