20
$\begingroup$

A 8x8 chess board is marked with the numbers from 1 to 64 sequentially from left to right, row by row and top to bottom. Now a minus sign is added to 32 of those numbers such that in each row and in each column there are exactly 4 positive and 4 negative numbers. All 64 numbers are now added up.

What is the smallest sum, which can be achieved?

enter image description here

$\endgroup$
  • 4
    $\begingroup$ Smallest as in smallest absolute value (closest to zero) or absolute smallest (most negative number)? $\endgroup$ – Rand al'Thor Feb 19 at 16:25
  • $\begingroup$ I see what you did there with the second tag. $\endgroup$ – Nautilus Feb 19 at 18:40
33
$\begingroup$

The smallest sum that can be achieved is

$0$

Because

This will always be the sum

Reasoning

The entries in the $n$th row in the given grid may be written as $$\{8n -7, 8n -6, 8n-5, 8n-4, 8n-3, 8n -2, 8n-1, 8n \} $$ Notice that when we negate four entries in a single row and take the row sum, the parts containing $n$ all cancel out. Since the total sum is just the sum of row sums, the problem is thus equivalent to one in which each row is $$\{1,2,3,4,5,6,7,8\}$$ But, in this version of the problem, each column contains the same entry, so if we negate four entries in each column then the sum of the elements in each column is $0$ and hence, the overall sum is $0$

| improve this answer | |
$\endgroup$
3
$\begingroup$

This is one of the first on puzzling.SE that I can actually do! :)

There are many ways to get $0$. Note that each row has precise $4$ or each $+/-$, so we can translate so that it's just eight copies of $\{1, ..., 8\}$. By symmetry, we have $$ 1 + 8 - 2 - 7 = 0 \quad\text{and}\quad 3+6 - 4-5 = 0.$$ So we can easily make each row sum to $0$. Other combinations work. Simply alternate these down each row, eg \begin{matrix} 1 & -2 & -3 & 4 & 5 & -6 & -7 & 8 \\ -1 & 2 & 3 & -4 & -5 & 6 & 7 & -8 \\ 1 & -2 & -3 & 4 & 5 & -6 & -7 & 8 \\ -1 & 2 & 3 & -4 & -5 & 6 & 7 & -8 \\ 1 & -2 & -3 & 4 & 5 & -6 & -7 & 8 \\ -1 & 2 & 3 & -4 & -5 & 6 & 7 & -8 \\ 1 & -2 & -3 & 4 & 5 & -6 & -7 & 8 \\ -1 & 2 & 3 & -4 & -5 & 6 & 7 & -8 \end{matrix} There are loads of similar combinations.

| improve this answer | |
$\endgroup$
1
$\begingroup$

If the summation cannot be a negative number, the minimum sum that can be achieved is 0. If the summation can be a negative number, then, the minimum that can be achieved is -107 (if the greatest numbers in each row is negated)

| improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ Each column must have 4 +s and 4 -s though too, so your latter suggestion doesn't fit. $\endgroup$ – Ben Reiniger Feb 20 at 20:38
  • $\begingroup$ @BenReinger: Say row 1 has 1,2,3,4,5,6,7,8 - the lowest sum can be achieved if the numbers are 1,2,3,4,-5,-6,-7,-8.. Similarly if all the largest numbers within every row are made negative numbers, the smallest sum is achieved. Which is -107. $\endgroup$ – Manoj Balu Feb 20 at 20:47
  • $\begingroup$ @ManojBalu - remember you're not adding a row, but all 64 numbers $\endgroup$ – SeanC Feb 20 at 21:15
  • 2
    $\begingroup$ @ManojBalu more importantly (I think) remember there has to be 4(+) and 4(-) in each column besides each row $\endgroup$ – TCooper Feb 21 at 1:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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