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Topic:
One Through Nine, With Eight Missing, And Two Ones
Replies:
19
Last Post:
Mar 31, 2011 5:11 PM




Re: One Through Nine, With Eight Missing, And Two Ones
Posted:
Mar 24, 2011 12:40 PM


On 3/24/11 12:19 PM, Ilan Mayer wrote: > On Mar 24, 10:54 am, Leroy Quet<qqq...@mindspring.com> wrote: >> Leroy Quet wrote: >>> ... >>> (The number that is the sum in any particular 3square/cell row, >>> column, or main diagonal may be in the square/cell on either end or >>> may be the central square/cell.) >> >>> ... >> >> This is confusing. I am talking about, in any particular row, column, >> or main diagonal, the largest integer (the sum) may be any one of the >> 3 integers of that particular row, column, or main diagonal. >> >> Thanks, >> Leroy Quet > > SPOILER > > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > . > > 312 > 167 > 459
I got the same answer, but with some reasoning. I mostly rely on the fact that there are only three triples containing 9: {2,7,9}, {3,6,9}, and {4,5,9}.
There are four lines through the center, so the 9 can't be in the center.
If the 9 is on a side, the other two numbers on that side form one of the three triples, then there are four possibilities for the center. In those twelve cases, fill in lines with the sum or difference to reach a quick contradiction.
If the 9 is on the corner, the diagonal not containing that corner must have one each of {2,7}, {3,6}, and {4,5}, so must be either {2,3,5}, {2,4,6}, or {3,4,7}, then in each case we have three possibilities for the center. The three triples then determine all the numbers in a line with 9, and the other two numbers must be 1. The above solution is the only one that works.
Is there a solution method that involves fewer than these 21 cases?
Dan



