By David Lewis

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Leonhard Euler created such a magic square with a constant of 260; stopping halfway on each horizontal or vertical line gives 130. Henry Dudeney created a magnificent square. Not only can the knight jump from the 64 to the 1 to restart its tour, but the diagonals are also nearly correct-256 and 264. 54 EUREKA! These large squares are not as hard to create as you might think. There is one very interesting method that produces a concentric, or bordered, magic square. These squares stay magic as their outside rows are repeatedly dissolved until the core of an order-3 or order-4 square is reached.

Otherwise, perform U'RU 2 D 2 L'. Go back to step 1. 6. If the cubie is in the bottom layer, turn it to fd. If the orange is on the bottom, perform RL'F 2 LR'. Otherwise, perform F'DU'R and go back to step 1. 7. Congratulate yourself. You've solved a layer-more than one-third of the cube! Equator Centers Turn the middle layer so that the centers match the trimmings of the top layer, as in Fig. 5-3. Fig. 5-3 65 Bottom Comers-Position 1. Turn the cube so that the orange is on the bottom. Choose a corner on top and correctly position it by turning Us several times so that its colors match the equator centers.

These are rather easy to devise, and you can no doubt create grander illusions yourself. 3. A number such as 141,414 is divisible by 10,101. As 10,101 is divisible by 7 and 13, 141,414 is, too. A number such as 1414 is divisible by 101; as 101 is prime, this number must be divisible only by 101 and 1. 4. Merely note the fourth number from the bottom, here 50. Multiplying this by 11 gives the sum. Or multiply the first number by 55 and the second by 88 and add the products. 5. The sum of the top faces is 11.