Author |
Topic: A summation exercise (Read 608 times) |
|
BNC
Uberpuzzler
Gender:
Posts: 1732
|
|
A summation exercise
« on: Feb 19th, 2003, 10:10am » |
Quote Modify
|
Complete the following summation exercise, using all digits 1 - 9 once. There are multiple answers. The one required is the one that yields the maximal sum. And please -- no brute force __ __ __ + __ __ __ ____________ __ __ __
|
|
IP Logged |
How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
|
|
|
poseur
Guest
|
|
Re: A summation exercise
« Reply #1 on: Feb 19th, 2003, 12:04pm » |
Quote Modify
Remove
|
By "no brute force" I assume you're not only ruling out computerized solutions but also just guessing till you find an answer. So here's what I was able to logic out. With 4 even digits and 5 odd divided into 3 sets, it's easy to see the answer will have to involve one instance of carrying a one to the next column. Assuming there will be at least one solution greater than 900; if tens are going to carry over to the hundreds digit, the highest total would be in the 950's, and it takes very little shuffling of the numbers to find that solution: 271+683=954. If it's the ones column that is>10, then let's assume we want to minimize the high numbers wasted in that column. 5,6,1. That easily points to the solution 235+746=981. In this scenario the tens and hundreds addends must total 16, so there's no way to increase the ones total.
|
|
IP Logged |
|
|
|
James Fingas
Uberpuzzler
Gender:
Posts: 949
|
|
Re: A summation exercise
« Reply #2 on: Feb 20th, 2003, 12:51pm » |
Quote Modify
|
Here's another way to get the same result: 1) With no carries, the sum of the six digits on top must equal the sum of the three digits on bottom. But the sum of all nine digits is 45, so there's no way to do this. 2) With one carry, we have the top numbers sum to 9 greater than the bottom numbers. (45-9)/2=18, so the top numbers sum to 27 and the bottom numbers sum to 18. 3) The biggest number we can make with three digits that are all different, yet sum to 18, is 981. 4) The carry cannot be in the hundreds digit, and it also can't be in the tens digit. Therefore, the carry is in the ones digit. A little trial-and-error reveals that you actually can form the sum 981, just as poseur found (or in a couple other ways).
|
|
IP Logged |
Doc, I'm addicted to advice! What should I do?
|
|
|
|