Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Problem New Solution

I believe I have gotten all of the ways now - thanks for the hints below Yun, Andre Nicolas, and Gerry Myerson. If anyone could confirm my answer (I feel there should be more possibilities, but for the numbers to be increasing left-right, I have found them all I believe). Sorry I don't have the headings on this chart, but from left to right it is: Numbers 1-6 (one for each column) with the last column being "Sum," if it was not already obvious.

share|improve this question
    
I'd suggest looking at the 16-example in the problem statement to see what's wrong with your attempts at the 26-problem. –  Gerry Myerson Feb 20 '12 at 4:40
    
I did, but I am still unsure of their methodology exactly. I edited my post to resemble somewhat of their "strategy." I edited my post with a new chart using the numbers from 16-12 so far, but I am still confused on how they are using their systematic list. –  Joe Feb 20 '12 at 4:46
    
I wuld probably divide everything by $2$, and look for $6$ non-zero numbers that add up to $13$. List say in non-decreasing order, like in the sample. You have $6$ numbers that add up to $26$, so the biggest has to be at least $6$. –  André Nicolas Feb 20 '12 at 6:19
    
The last three rows of your most recent chart differ only in the order of the numbers, so it's evident from the 16-example that they don't count as different. In fact, it's evident from the 16-example that you are allowed to assume that the numbers in a row never get smaller as you go from left to right. That will make it a lot easier to list all the possibilities, if you choose to do that. –  Gerry Myerson Feb 20 '12 at 10:13
    
Thanks for the help. I believe I have reached a solution. If anyone could verify my solution or see if there are "gaps" in my solution, that would be appreciated. Again, I believe I have gotten all of the ways if the numbers increase from left to right. –  Joe Feb 20 '12 at 17:59
add comment

1 Answer

up vote 2 down vote accepted

I can verify your solution. I have one more than you. I've also put up a list (everything is divided by 2). I also put it up in the order that mirrors the smaller example's list (at least, as I interpreted it - perhaps, putting the list in strictly number-increasing order would have been a good idea too).

1,1,1,1,1,8

1,1,1,1,2,7

1,1,1,1,3,6

1,1,1,2,2,6

1,1,1,1,4,5

1,1,1,2,3,5

1,1,2,2,2,5

1,1,1,2,4,4

1,1,1,3,3,4

1,1,2,2,3,4

1,2,2,2,2,4

1,1,2,3,3,3

1,2,2,2,3,3

2,2,2,2,2,3

share|improve this answer
    
Many thanks mixedmath. Our lists are very similar in terms of the set-up. I simply missed that one way of "1,2,2,2,2,4" on your list - thanks for the catch. –  Joe Feb 20 '12 at 19:02
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.