Guest Posted August 23, 2011 Report Share Posted August 23, 2011 using only numbers 1-9 in a 3 by 3 grid. You use all nine numbers only once so that each column is a correct addition going down. We have not been able to solve it. Maybe there is no solution. Please help me out. Quote Link to comment Share on other sites More sharing options...
0 fabpig Posted August 23, 2011 Report Share Posted August 23, 2011 2 9 4 7 5 3 6 1 8 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 23, 2011 Report Share Posted August 23, 2011 do you mean so that each column and row adds up to the same number? if so then i think ive figured it out.... 1 6 8 =15 5 7 3 =15 9 2 4 =15 = = = 15 15 15 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 23, 2011 Report Share Posted August 23, 2011 If you mean so that A+B=C for each column in a grid like this: A A A B B B C C C Then it's not possible. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 23, 2011 Report Share Posted August 23, 2011 As I read the question, you are looking for something like A B C D E F G H I where A + D = G, B + E = H, C + F = I and [A - I] span 1-9 completely. There is no solution to this problem as shown by the following argument. We have 5 odd numbers and 4 even numbers to use. Assume that G, H and I are all odd, then you will need three more odd numbers in [A-F]. But that would make a total of 6 odds and you had 5 to start with. Assume that two of the sums are odd and the third is even. Then we need two odd numbers to get the odd sums--for a total of 4 odd numbers used. But we have a fifth odd number to use and the final sum is even which creates another contradiction. Assume that only one one sum is odd. Then you need to have either 1, 3 or 5 odd numbers to get those sums for a total of 2, 4 or 6 odd numbers (another contradiction). Assuming that none of the sums is odd, then you need to have started with an even number of odd numbers. So it can not be done. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 23, 2011 Report Share Posted August 23, 2011 First we have to know what the sum of the nine integers is equal to. We know that the sum of the first n digits is: (n*n+1)/2 Hence, the sum of the first 9 digits is: 9*10/2 = 45 We divide 45 by 3 = 15. This is the number that each of the columns should sum. This fact limits your combinations for the trial and error very much and can come up with an answer really fast: 123 564 978 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 23, 2011 Report Share Posted August 23, 2011 (edited) I read the problem to mean each column makes a correct addition problem (i.e. a column could be 1, 3, 4, or 1+3=4) In that case, here's one answer 416 273 689 Edited August 23, 2011 by bobbyplump Quote Link to comment Share on other sites More sharing options...
0 Guest Posted August 23, 2011 Report Share Posted August 23, 2011 And I just realized I'm an idiot. Quote Link to comment Share on other sites More sharing options...
0 Smith Posted August 23, 2011 Report Share Posted August 23, 2011 And I just realized I'm an idiot. You're no idiot for making a mistake! People who never take a chance rarely succeed. You understood the question, formulated a proper response, but missed one little detail... It's OK to be wrong as long as you accept that you are wrong when it's pointed out, or don't try to cover it up when you see your mistake. Quote Link to comment Share on other sites More sharing options...
0 Smith Posted August 23, 2011 Report Share Posted August 23, 2011 As I read the question, you are looking for something like A B C D E F G H I where A + D = G, B + E = H, C + F = I and [A - I] span 1-9 completely. There is no solution to this problem as shown by the following argument. We have 5 odd numbers and 4 even numbers to use. Assume that G, H and I are all odd, then you will need three more odd numbers in [A-F]. But that would make a total of 6 odds and you had 5 to start with. Assume that two of the sums are odd and the third is even. Then we need two odd numbers to get the odd sums--for a total of 4 odd numbers used. But we have a fifth odd number to use and the final sum is even which creates another contradiction. Assume that only one one sum is odd. Then you need to have either 1, 3 or 5 odd numbers to get those sums for a total of 2, 4 or 6 odd numbers (another contradiction). Assuming that none of the sums is odd, then you need to have started with an even number of odd numbers. So it can not be done. I believe TreesFearMe understood the question correctly, provided the correct answer, and an excellent proof! On the other hand, I am NOT the OP'er, so what do I know? Quote Link to comment Share on other sites More sharing options...
0 Molly Mae Posted August 23, 2011 Report Share Posted August 23, 2011 I believe TreesFearMe understood the question correctly, provided the correct answer, and an excellent proof! On the other hand, I am NOT the OP'er, so what do I know? Hey, Smith. It's been a bit. I agree with you and Trees. Consider using the range 0-8 instead. Where's the immediate flaw? Quote Link to comment Share on other sites More sharing options...
0 superprismatic Posted August 23, 2011 Report Share Posted August 23, 2011 Funny, but you can't even do it with a 4 by 4 grid using the numbers 1-16, or even 0-15! Quote Link to comment Share on other sites More sharing options...
0 Molly Mae Posted August 25, 2011 Report Share Posted August 25, 2011 (edited) Funny, but you can't even do it with a 4 by 4 grid using the numbers 1-16, or even 0-15! I got it using a 1x3 grid using 1-3. =P But not 0-2. Can a 3x3 grid using 1-9 be completed if you allow for multiplication and addition? How many unique solutions are there? 4x4 using 1-16? Using 0-15? Edited August 25, 2011 by Molly Mae Quote Link to comment Share on other sites More sharing options...
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using only numbers 1-9 in a 3 by 3 grid. You use all nine numbers only once so that each column is a correct addition going down. We have not been able to solve it. Maybe there is no solution. Please help me out.
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