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Magic square help needed

janvianish

New member
Jun 26, 2012
3
HI

In a square 3 x 3 using the numbers 1 to 9 once only put the numbers so that:
the numbers on the top row minus the numbers in the 2nd row = the numbers on the 3rd row.

trying this for about 4 hrs and am always 1 number out.
 

Unknown008

Member
Feb 1, 2012
20
Um.. one possible answer:

846, 327 and 519

It was rather trial and error though, and departing from that I should be getting the first digits to be subtracted requiring borrowing from the tens column.
 

janvianish

New member
Jun 26, 2012
3
hi thanks for this but i do not think we can borrow from another column
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Unknown008's neat trick of borrowing from the tens column is the only way to solve this problem. If borrowing is not allowed then the problem has no solution.

In fact, each number in the top row would then have to be the sum of the two numbers below it. But an odd number must be the sum of an odd number and an even number; and an even number is either the sum of two even numbers or the sum of two odd numbers. That means that each column must contain either two odd numbers or no odd numbers. Therefore the total number of odd numbers in the square must be even. But there are five odd numbers in the set 1, ..., 9, and five is not an even number. So there is no possibility to fill the square in the required way (except by using the borrowing trick).
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,502
If borrowing is allowed, then exhaustive search shows that there are 336 solutions.
 

janvianish

New member
Jun 26, 2012
3
thanks to all for your help....