Numbers by Alfred S. Posamentier

Author:Alfred S. Posamentier
Language: eng
Format: epub
ISBN: 9781633880313
Publisher: Prometheus Books
Published: 2015-07-17T04:00:00+00:00

Figure 7.9: A map of a general magic square

In a magic square, we would thus have

r2 + c2 + d1 + d2 = 15 + 15 + 15 + 15 = 60.

However, this sum can also be written as

r2 + c2 + d1 + d2 = (d + e + f) + (b + e + h) + (a + e + i) + (c + e + g) =

3 e + (a + b + c + d + e + f + g + h + i) = 3e + 45

Therefore, 3e + 45 = 60, and e = 5. Thus it is established that the center position of a magic square of order 3 must be occupied by the number 5.

Recall that two numbers of an nth order magic square are said to be complementary if their sum is n2 + 1. In a 3 × 3 magic square, two numbers are complementary if their sum is 9 + 1 = 10. We can now see that numbers on opposite sides of 5 are complementary. For example, a + i = d1 – e = 15 – 5 = 10, and, therefore, a and i are complementary. But so are the pairs g and c, b and h, and d and f.

Let us now try to put 1 in a corner, as shown in figure 7.10. Here a = l, and therefore i must be 9, so that the diagonal adds up to 15. Next we notice that 2, 3, and 4 cannot be in the same row (or column) as 1, since there is no natural number less than 9 that would be large enough to occupy the third position of such a row (or column). This would leave only the two shaded positions in figure 7.10 to accommodate these three numbers (2, 3, and 4). Since this cannot be the case, our first attempt was a failure: the numbers 1 and 9 may occupy only the middle positions of a row (or column).

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