Magic sums

Logic Level 2

Consider all magic squares of size 3 which use non-repeated positive integers.

Which of the following options could be the magic sum?

14 12 18 16 13

1 solution

Lc Nemo
Dec 11, 2017

The smallest magic square that contains non-repeated positive integers contain the integers 1-9, and such a magic square gives the magic sum of 15. Any other magic square that contains non-repeated positive integers will have to replace any of the integers 1 to 9 for one of a higher value x>9. This will lead to a magic sum >15. The only option that is greater than 15 is the only possible option.

Thanks. I've added 16 as an option. Can you add more details to explain why 18 is the next possible magic sum?

Calvin Lin Staff - 3 years, 5 months ago

Nice. You've plugged a loophole. I honestly didn't know how to solve for possible magic sums.

Maintaining the criteria of non-repeated positive integers, yielding the smallest possible magic sum = 15 as explained above, the next magic sum is obtained by adding 1 to each integer in the original magic square. We need to add 1 to every integer for two reasons: to keep their uniqueness (i.e. none of them repeats). Because, if we add 1 to 1, but keep 2, we end up with 2 appearing twice. And to keep the magic sum. Because, if we add 1 to any integer, this will raise the sum of the row and column which it occupies by 1, but not the other rows and columns, causing a difference, and magic sums are supposed to be equal.

This will give us integers from 2 to 10, a total sum of 54, and a magic sum of 18.

The next larger magic sum, according to my clumsy explanation, is thus 21.

Generalising, we get the set of magic sums = { x is an integer | x > 14 and x = 3a where a is an integer}

Lc Nemo - 3 years, 5 months ago

More accurately, if a magic square exists which uses the integers from $n$ to $n+8$ , then the sum of all 3 rows will be $9n+36$ , and hence the magic sum will be $3n + 12$ . Conversely, we have to show that a magic square exists for these integers, which follows by just adding $n-1$ to our usual magic square.

Calvin Lin Staff - 3 years, 5 months ago

@Calvin Lin – Thanks! You've summed it up elegantly.

Lc Nemo - 3 years, 5 months ago

Magic sums

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